assignmnet 11 qa

course Phy 201

011. Note that there are 12 questions in this set.

.Situations involving forces and accelerations.

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Question: `q001. A cart on a level, frictionless surface contains ten masses, each of mass 2 kg. The cart itself has a mass of 30 kg. A light weight hanger is attached to the cart by a light but strong rope and suspended over the light frictionless pulley at the end of the level surface.

Ignoring the weights of the rope, hanger and pulley, what will be the acceleration of the cart if one of the 2 kg masses is transferred from the cart to the hanger?

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Your solution: Fnet=m*g=2kg *9.8=19.6N

This is the net force that would act on the whole system to cause acceleration.

The total mass of the system=30+10*2=50kg

F=m*a=50*a=19.6

a=.392m/sec square.

Confidence rating: Pretty confident after watching the video on atwood machine.

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Given Solution:

At the surface of the Earth gravity, which is observed here near the surface to accelerate freely falling objects at 9.8 m/s^2, must exert a force of 2 kg * 9.8 m/s^2 = 19.6 Newtons on the hanging 2 kg mass.

This force will tend to accelerate the system consisting of the cart, rope, hanger and suspended mass, in the 'forward' direction--the direction in which the various components of the system must accelerate if the hanging mass is to accelerate in the downward direction. This is the only force accelerating the system in this direction.

All other forces, including the force of gravity pulling the cart and the remaining masses downward and the normal force exerted by the level surface to prevent gravity from accelerating the cart downward, are in a direction perpendicular to the motion of these components of the system; furthermore these forces are balanced so that they add up to 0.

The mass of the system is that of the 30 kg cart plus that of the ten 2 kg masses, a total of 50 kg.

The net force of 19.6 Newtons exerted on a 50 kg mass therefore results in acceleration

a = Fnet / m = 19.6 Newtons / (50 kg)

= 19.6 kg m/s^2 / (50 kg)

= .392 m/s^2.

STUDENT QUESTION:

Our answers are close, but wouldn’t the 50kg actually by 48kg since the 2kg weight was

taken out of the kart?

INSTRUCTOR RESPONSE:

The 2 kg weight is still part of the mass that's being accelerated. It's been moved from the cart to the hanger, but it's still there. All 50 kg are being accelerated by that net force.

STUDENT QUESTION

Am I allowed to divide Newtons by Kg? Or do I have to change the Newtons to kg*m/s^2?

INSTRUCTOR RESPONSE

You need to reduce everything to fundamental units. How would you know what N / kg is unless they are both expressed in compatible units?

You could of course memorize the fact that N / kg gives you m/s^2, along with about 200 other shortcuts, but that would be a waste of time and wouldn't contribute much to your insight or your ability to work out units in unfamiliar situations.

You can count the number of fundamental units in all of physics on one hand. If you know the definitions of the quantities, this makes it very easy to deal with questions of units. Unit calculations come down to the simple algebra of multiplying and dividing fractions, whose numerators and denominators are just products and powers of a few simple units.

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Self-critique (if necessary):

Self-critique rating:

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Question: `q002. Two 1-kg masses are suspended over a pulley, one on either side. A 100-gram mass is added to the 1-kg mass on one side of the pulley. How much force does gravity exert on each side of the pulley, and what is the net force acting on the entire 2.1 kg system?

Your solution: Force of gravity on 1kg side=9.8*1=9.8N

Force of gravity on 1.1 kg side=9.8*1.1= 10.78N

Net force would be equal to downward force –upward force.

Heavier weight/force would exert a downward force

So, 10.78-9.8=.98N as the net force.

Confidence rating: Very confident

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Given Solution:

The 1-kg mass experiences a force of F = 1 kg * 9.8 m/s^2 = 9.8 Newtons. The other side has a total mass of 1 kg + 100 grams = 1 kg + .1 kg = 1.1 kg, so it experiences a force of F = 1.1 kg * 9.8 Newtons = 10.78 Newtons.

Both of these forces are downward, so it might seem that the net force on the system would be 9.8 Newtons + 10.78 Newtons = 20.58 Newtons. However this doesn't seem quite right, because when one mass is pulled down the other is pulled up so in some sense the forces are opposing. It also doesn't make sense because if we had a 2.1 kg system with a net force of 20.58 Newtons its acceleration would be 9.8 m/s^2, since we know very well that two nearly equal masses suspended over a pulley won't both accelerate downward at the acceleration of gravity.

So in this case we take note of the fact that the two forces are indeed opposing, with one tending to pull the system in one direction and the other in the opposite direction.

We also see that we have to abandon the notion that the appropriate directions for motion of the system are 'up' and 'down'. We instead take the positive direction to be the direction in which the system moves when the 1.1 kg mass descends.

We now see that the net force in the positive direction is 10.78 Newtons and that a force of 9.8 Newtons acts in the negative direction, so that the net force on the system is 10.78 Newtons - 9.8 Newtons =.98 Newtons.

The net force of .98 Newtons on a system whose total mass is 2.1 kg results in an acceleration of .98 Newtons / (2.1 kg) = .45 m/s^2, approx.. Thus the system accelerates in the direction of the 1.1 kg mass at .45 m/s^2.

Additional note on + and - directions:

One force tends to accelerate the system in one direction, the other tends to accelerate it in the opposite direction.

• So you need to choose a positive direction and put a + or - sign on each force, consistent with your chosen positive direction.

The positive direction can't be 'up' or 'down', since part of the system moves up while another part moves down.

• The easiest way to specify a positive direction is to specify the direction of one of the masses.

Self-critique (if necessary)

Self-critique rating:

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Question: `q003. If in the previous question there is friction in the pulley, as there must be in any real-world pulley, the system in the previous problem will not accelerate at the rate calculated there. Suppose that the pulley exerts a retarding frictional force on the system which is equal in magnitude to 1% of the weight of the system. In this case what will be the acceleration of the system, assuming that it is moving in the positive direction (as defined in the previous exercise)?

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Your solution: Acceleration of the frictionless pulley system=

F=m*a

.98=2.1a

a=.45m/sec square

Frictional force on the system= 1% of weight of system

= 1% (2.1*9.8)

= 1% (20.58)=.21 N

Since this is the retarding force acting in opposite direction,this will be negative force=-.21N

The net force acting on this system=10.78 N - 9.8 N-.21N=.77N

Acceleration=Fnet/mass=.77/2.1=.37m/sec square

Confidence rating: Very confident

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Given Solution:

We first determine the force exerted a friction. The weight of the system is the force exerted by gravity on the mass of the system. The system has mass 2.1 kg, so the weight of the system must be

2.1 kg * 9.8 m/s^2 = 20.58 Newtons.

1% of this weight is .21 Newtons, rounded off to two significant figures. This force will be exerted in the direction opposite to that of the motion of the system; since the system is assume to be moving in the positive direction the force exerted by friction will be

frictional force = -.21 Newtons.

The net force exerted by the system will in this case be 10.78 Newtons - 9.8 Newtons - .21 Newtons = .77 Newtons, in contrast with the .98 Newton net force of the original exercise.

The acceleration of the system will be

.77 Newtons / (2.1 kg) = .37 m/s^2, approx..

STUDENT SOLUTION

(this solution and the instructor's commentary address a common error in expression and in thinking

this is worth a look

the main topic is why it's not appropriate to write an expression like .98N-.21N=.77N/2.1kg=.37m/s^2

Using my numbers from the previous problem, we had .98N pulling down, so I will subtract an additional .21

from that to get the Newtons pulling down w/ friction. .98N-.21N=.77N/2.1kg=.37m/s^2.

INSTRUCTOR COMMENTARY

It's clear what you mean by .98N-.21N=.77N/2.1kg=.37m/s^2, and everything you said up to this point is very good and correct.

However as a mathematical statement .98N-.21N=.77N/2.1kg=.37m/s^2 is incorrect.

If .98N-.21N=.77N/2.1kg=.37m/s^2, then since quantities that are both equal to a third quantity are equal to one another, .98N-.21N = .37m/s^2.

However N and m/s^2 are complete different units, so the left- and right-hand sides of this equality are unlike terms. Unlike terms can't possibly be equal.

And of course it's very obvious that

.98N-.21N =. .37m/s^2

is simply an untrue statement.

Untrue statements tend to lead to confusion, e.g., when you review your work and don't necessarily remember what you were thinking when you wrote the thing down, or when your statement is viewed by someone else who doesn't already know what to expect.

If you said

F_net = .98N-.21N=.77N

so

a = F_net / m = .77 N / (2.1kg) =.37m/s^2

then you would not have any false statements in your solution, your solution would be clear to anyone who understands Newton's Second Law, and would be much more likely useful to you when reviewing your work.

(Note also that N / kg = (kg m/s^2) / kg = m/s^2

It's important to maintain the habit of reducing units to fundamental units and doing the algebra of the units.)

Self-critique (if necessary

Self-critique rating:

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Question: `q004. Why was it necessary in the previous version of the exercise to specify that the system was moving in the direction of the 1.1 kg mass. Doesn't the system have to move in that direction?

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Your solution: Well, this is exactly like in the experiment video I saw.

When the professor gave a little nudge, even the less weight/force side did go downward. So that means, by that nudge professor was adding initial velocity to the system.Therfore, not always will the system move downward on the heavier weight side. It may be different if initial velocity is added to the system

Confidence rating: Very confident as I just saw the professor’s experiment video.

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Given Solution:

If the system is released from rest, since acceleration is in the direction of the 1.1 kg mass its velocity will certainly always be positive.

However, the system doesn't have to be released from rest. We could give a push in the negative direction before releasing it, in which case it would continue moving in the negative direction until the positive acceleration brought it to rest for an instant, after which it would begin moving faster and faster in the positive direction.

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Self-critique (if necessary):I need to remember the concept that the positive acceleration would bring the system to rest for a minute

Self-critique rating:

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Question: `q005. If the system of the preceding series of exercises is initially moving in the negative direction, then including friction in the calculation what is its acceleration?

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Your solution: It’s easy to calculate the acceleration, but I am not sure if we will reverse the direction of all the forces acting on the system or just the friction.

I mean

Net force=10.78 N -9.8 N+.21N

OR -10.78 N +9.8N+.21N

Confidence rating: Doubt ful about the two possibilities I mentioned above

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Given Solution:

Its acceleration will be due to the net force. This net force will include the 10.78 Newton force in the positive direction and the 9.8 Newton force in the negative direction. It will also include a frictional force of .21 Newtons in the direction opposed to motion.

Since motion is in the negative direction, the frictional force will therefore be in the positive direction. The net force will thus be

Fnet = 10.78 Newtons - 9.8 Newtons + .21 Newtons = +1.19 Newtons,

in contrast to the +.98 Newtons obtained when friction was neglected and the +.77 Newtons obtained when the system was moving in the positive direction. ,

The acceleration of the system is therefore

• a = 1.19 N / (2.1 kg) = .57 m/s^2.

Self-critique (if necessary)

Self-critique rating:

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Question: `q006. If friction is neglected, what will be the result of adding 100 grams to a similar system which originally consists of two 10-kg masses, rather than the two 1-kg masses in the previous examples?

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Your solution: I am just guessing without really solving the whole problem that the net force should not change as the mass is increased to 10 times in both the directions.

Acceleration can be calculated by dividing the net force by the total mass of the system which in this case would be 20.1kg

Confidence rating: Have not solved the whole question. But I know it’s the same method like we did in the first few questions

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Given Solution:

In this case the masses will be 10.1 kg and 10 kg. The force on the 10.1 kg mass will be 10.1 kg * 9.8 m/s^2 = 98.98 Newtons and the force on the 10 kg mass will be 10 kg * 9.8 m/s^2 = 98 Newtons.

The net force will therefore be .98 Newtons, as it was in the previous example where friction was neglected.

We note that this.98 Newtons is the result of the additional 100 gram mass, which is the same in both examples.

The total mass of the system is 10 kg + 10.1 kg = 20.1 kg, so that the acceleration of the system is

• a = .98 Newtons / 20.1 kg = .048 m/s^2, approx..

Comparing this with the preceding situation, where the net force was the same (.98 N) but the total mass was 2.1 kg, we see that the same net force acting on the significantly greater mass results in significantly less acceleration.

Note on the direction of the frictional force: It's not quite accurate to say that the frictional force is always in the direction opposite motion. I'm not really telling you the whole story here--trying to keep things simple. Friction can indeed speed things up, depending on your frame of reference.

The more accurate statement is that forces exerted by kinetic friction act in the direction opposite the relative motion of the two surfaces. (Forces exerted by static friction act in the direction opposite the sum of all other forces).

• For example a concrete block, free to slide around in the bed of a pickup truck which is accelerating forward, is accelerated by the frictional force between it and the truckbed. So the frictional force is in its direction of motion. If the block doesn't slide, it is static friction that accelerates it and there is no relative motion between the surfaces of the block and the truckbed. If the block does slide, the frictional force is still pushing it forward relative to the road, and relative to the road it accelerates in its direction of motion, but the frictional force isn't sufficient to accelerate it at the same rate as the truck; it therefore slides backward relative to the truckbed. Relative to the truckbed the block slides backward while the frictional force pushes it forward--the frictional force is in the direction opposite the relative motion.

• If the block is sliding, it is moving toward the back of the truck while friction is pushing it toward the front. So in this case the frictional force acts in the direction opposite the relative motion of the two surfaces.

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Self-critique (if necessary):I need to remember that friction is not always in direction opposite to motion rather forces exerted by kinetic friction act in the direction opposite the relative motion of the two surfaces.

Self-critique rating:

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Question: `q007. If friction is not neglected, what will be the result for the system with the two 10-kg masses with .1 kg added to one side? Note that by following what has gone before you could, with no error and through no fault of your own, possibly get an absurd result here, which will be repeated in the explanation then resolved at the end of the explanation.

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Your solution: Its solved in the same way as earlier.

Confidence rating: I am sorry for not solving the whole problem but I know that I do understand the concept very well.

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Given Solution:

If friction is still equal to 1% of the total weight of the system, which in this case is 20.1 kg * 9.8 m/s^2 = 197 Newtons, then the frictional force will be .01 * 197 Newtons = 1.97 Newtons. This frictional force will oppose the motion of the system.

For the moment assume the motion of the system to be in the positive direction. This will result in a frictional force of -1.97 Newtons. The net force on the system is therefore 98.98 Newtons - 98 Newtons - 1.97 Newtons = -.99 Newtons.

This net force is in the negative direction, opposite to the direction of the net gravitational force. If the system is moving this is perfectly all right--the frictional force being greater in magnitude than the net gravitational force, the system can slow down.

Suppose the system is released from rest. Then we might expect that as a result of the greater weight on the positive side it will begin accelerating in the positive direction. However, if it moves at all the frictional force would result in a -.99 Newton net force, which would accelerate it in the negative direction and very quickly cause motion in that direction. Of course friction can't do this--its force is always exerted in a direction opposite to that of motion--so friction merely exerts just enough force to keep the object from moving at all.

Friction acts as though it is quite willing to exert any force up to 1.97 Newtons to oppose motion, and up to this limit the frictional force can be used to keep motion from beginning.

In fact, the force that friction can exert to keep motion from beginning is usually greater than the force it exerts to oppose motion once it is started.

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Self-critique (if necessary):

Self-critique rating:

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Question: `q008. A cart on an incline is subject to the force of gravity. Depending on the incline, some of the force of gravity is balanced by the incline. On a horizontal surface, the force of gravity is completely balanced by the upward force exerted by the incline. If the incline has a nonzero slope, the gravitational force (i.e., the weight of the object) can be thought of as having two components, one parallel to the incline and one perpendicular to the incline. The incline exerts a force perpendicular to itself, and thereby balances the weight component perpendicular to the incline. The weight component parallel to the incline is not balanced, and tends to accelerate the object down the incline. Frictional forces tend to resist this parallel component of the weight and reduce or eliminate the acceleration.

A complete analysis of these forces is best done using the techniques of vectors, which will be encountered later in the course. For now you can safely assume that for small slopes (less than .1) the component of the gravitational force parallel to the incline is very close to the product of the slope and the weight of the object. [If you remember your trigonometry you might note that the exact value of the parallel weight component is the product of the weight and the sine of the angle of the incline, that for small angles the sine of the angle is equal to the tangent of the angle, and that the tangent of the angle of the incline is the slope. The product of slope and weight is therefore a good approximation for small angles or small slopes.]

What therefore would be the component of the gravitational force acting parallel to an incline with slope .07 on a cart of mass 3 kg?

Your solution

As mentioned above, the component of gravitation force acting parallel to an incline=slope* weight

.07*(3*9.8)=2.058

Confidence rating:

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Given Solution:

The gravitational force on a 3 kg object is its weight and is equal to 3 kg * 9.8 m/s^2 = 29.4 Newtons. The weight component parallel to the incline is found approximately as (parallel weight component) = slope * weight = .07 * 29.4 Newtons = 2.1 Newtons (approx.).. STUDENT COMMENT: It's hard to think of using the acceleration of 9.8 m/s^2 in a situation where the object is not free falling. Is weight known as a force measured in Newtons? Once again I'm not used to using mass and weight differently. If I set a 3 kg object on a scale, it looks to me like the weight is 3 kg.INSTRUCTOR RESPONSE: kg is commonly used as if it is a unit of force, but it's not. Mass indicates resistance to acceleration, as in F = m a. {}{}eight is the force exerted by gravity. {}{}The weight of a given object changes as you move away from Earth and as you move into the proximity of other planets, stars, galaxies, etc.. As long as the object remains intact its mass remains the same, meaning it will require the same net force to give it a specified acceleration wherever it is.{}{}An object in free fall is subjected to the force of gravity and accelerates at 9.8 m/s^2. This tells us how much force gravity exerts on a given mass: F = mass * accel = mass * 9.8 m/s^2. This is the weight of the object. **

STUDENT NOTE:

i didn't think to use the acceleration of gravity for this one, it said the object was paralell

INSTRUCTOR RESPONSE:

The situation talks about the weight having two components, one being parallel to the incline, and the instruction tells you how to find that parallel component when the slope of the incline is small.

We know from experience that an object will pick up speed along the incline, as opposed to the direction perpendicular to the incline (to move in the perpendicular direction the object would have to leave the incline, either burrowing down into the incline or levitating up off the incline).

The direction along the incline is parallel to the incline. So its acceleration is parallel to the incline, and the net force must be parallel to the incline.

In the absence of other forces, only gravity has a component parallel to the incline. Therefore in this ideal case the gravitational component parallel to the incline is the net force. In reality there are other forces present (e.g., friction) but the parallel gravitational component is nevertheless present, and contributes to the net force in the direction of motion.

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Self-critique (if necessary):I need to remember that the parallel component of gravitational slope given by slope* weight is actually the net force

this would be the net force in a situation where there was no friction and no other external forces

Self-critique rating:

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Question: `q009. What will be the acceleration of the cart in the previous example, assuming that it is free to accelerate down the incline and those frictional forces are negligible?

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Your solution: Net force=parallel component = 2.058N

Mass =3kg

a=2.058/3=.686m/sec ^2

Confidence rating:

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Given Solution:

The weight component perpendicular to the incline is balanced by the perpendicular force exerted by the incline. The only remaining force is the parallel component of the weight, which is therefore the net force. The acceleration will therefore be a = F / m = 2.1 Newtons / (3 kg) = .7 m/s^2.

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Self-critique (if necessary):

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Question: `q010. What would be the acceleration of the cart in the previous example if friction exerted a force equal to 2% of the weight of the cart, assuming that the cart is moving down the incline? [Note that friction is in fact a percent of the perpendicular force exerted by the incline; however for small slopes the perpendicular force is very close to the total weight of the object].

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Your solution: Friction = 2% of 3*9.8=.588.This will be upwards causing the acceleration and therefore would be negative

Net force=2.058-.588=1.47N

a=1.47/3=.49m/s^2

Confidence rating: Very confident

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Given Solution:

The weight of the cart was found to be 29.4 Newtons. The frictional force will therefore be .02 * 29.4 Newtons = .59 Newtons approx.. This frictional force will oppose the motion of the cart, which is down the incline.

If the downward direction along the incline is taken as positive, the frictional force will be negative and the 2.1 Newton parallel component of the weight will be positive. The net force on the object will therefore be

net force = 2.1 Newtons - .59 Newtons = 1.5 Newtons (approx.).

This will result in an acceleration of

a = Fnet / m = 1.5 Newtons / 3 kg = .5 m/s^2.

Self-critique (if necessary)

Self-critique rating:

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Question: `q011. Given the conditions of the previous question, what would be the acceleration of the cart if it was moving up the incline?

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Your solution: The frictional force would not act downwards, therefore it will be positive.

But I am not sure if the direction of net force (the parallel component)would be reversed too ,

Confidence rating:

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Given Solution:

In this case the frictional force would still have magnitude .59 Newtons, but would be directed opposite to the motion, or down the incline. If the direction down the incline is still taken as positive, the net force must be

net force = 2.1 Newtons + .59 Newtons = 2.7 Newtons (approx).

The cart would then have acceleration

a = Fnet / m = 2.7 Newtons / 3 kg = .9 m/s^2.

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Self-critique (if necessary): I always get confused if I change in direction of motion is making any difference to net force. I understand that the direction of frictional force would be revered

Self-critique rating:

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Question: `q012. Assuming a very long incline, describe the motion of the cart which is given an initial velocity up the incline from a point a few meters up from the lower end of the incline. Be sure to include any acceleration experienced by the cart.

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Your solution:

Confidence rating:

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Given Solution:

The cart begins with a velocity up the incline, which we still taken to be the negative direction, and an acceleration of +.9 m/s^2. This positive acceleration tends to slow the cart while it is moving in the negative direction, and the cart slows by .9 m/s every second it spends moving up the incline.

Eventually its velocity will be 0 for an instant, immediately after which it begins moving down the incline as result of the acceleration provided by the weight component parallel to the incline.

As soon as it starts moving down the incline its acceleration decreases to +.5 m/s^2, but since the acceleration and velocity are now parallel the cart speeds up, increasing its velocity by .5 m/s every second, until it reaches the lower end of the incline.

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Self-critique (if necessary):

Self-critique rating:

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You're doing well with these problem. I'm glad you found the videos helpful.