course Phys 201
011. Note that there are 12 questions in this set.
.Situations involving forces and accelerations.
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?
m1=48kg
m2=2kg
m=m1+m2
=48kg+2kg
m=50kg
Fnet=2kg*9.8m/s^2
fNET= 19.6N
a=Fnet/m
a=19.6N/50kg
a=0.392m/s^2
Confidence Assessment:
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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.
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Self-critique (if necessary): I understand this question.
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?
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Your solution:
100g=.1kg
Fnet=1.1kg*9.8m/s^2
=10.18N
10.78N-9.8N=.98N
.98N/2.1kg=.46m/s^2 in the direction of 1.1kg
Confidence Assessment: 2
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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, and 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 to 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.
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Self-critique (if necessary): i understand this problem.
Self-critique rating:3
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:
2,1kg*9.8m/s^2=20.58N
20.59N*.01=.21N
Ffrct=-,21N
Fnet=F-m-Ffrict
=10.78N-9.8N-.21N
=.77N
a=Fnet/m
=.77N/2.1kg
=.37m/s^2
Confidence Assessment: 2
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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) = .35 m/s^2, approx..
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Self-critique (if necessary):i understand this problem.
Self-critique rating:3
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:
It is important to know the direction of the pulley because ti lets you know which way the direction the friction is going.
Confidence Assessment:
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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):
wouldn't it also be necessary to know the direction so you know the direction of the resiestive force and direction of gravity.
Confidence Assessment:
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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.
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Self-critique (if necessary):
Self-critique rating:
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?
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Your solution:
Confidence Assessment:
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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, and 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 to 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.
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Self-critique (if necessary):
Self-critique rating:
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:
Confidence Assessment:
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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) = .35 m/s^2, approx..
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Self-critique (if necessary):
Self-critique rating:
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:
Confidence Assessment:
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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):
Self-critique rating:
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:
Fnet=10.78N-9.8N+.21N
=1.19N
1.19N/2.1kg=.57m/s^2
Confidence Assessment: 3
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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.
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Self-critique (if necessary): i understand this problem.
Self-critique rating:3
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:
100g=.1kg
10.1kg*9.8m/s^2=98.98N
10*9.8m/s^2=98N
98.98N-98N=.98N
10.1+10=20.1kg
a=.98N/20.1kg
=.05m/s^2
Confidence Assessment:
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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 understand this problem.
Self-critique rating:
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:
&#Good responses. Let me know if you have questions. &#