@& v^2 would give you (m/s)^2 or m^2 / s^2. So this would be kg m^2 / s^2.*@

@& Acceleration is roc of vel wrt clock time, which is `dv / `dt.

`dv is 5 m/s, `dt is 12 s. The / 2 is not part of the correct calculation. I see how that happens; sort of a habit from when you average the two velocities to get average velocity.

`ds is the symbol for change in position, not acceleration.*@

@& Reasoning is right; acceleration should be double that so you would get double that force.*@

@& That's your initial KE.

You need to find final KE and subtract init from final.

*@

@& It's more than that. That's the initial KE.

*@

@& The acceleration of gravity is 10 m/s^2, but gravity doesn't act parallel to the incline. Only the component of gravity parallel to the incline contributes to the net force (the other component is balanced by the incline itself).

*@

@& You're right, you can't just think this one through.

You need to use the equations. You know `ds, a and v0.

Check the appended document for more discussion.*@

@& Again you need the equations.*@

@& `q001. The meter, the kilogram and the second are the basic SI units (SI stands for 'standard international').

Velocity is the rate of change of position with respect to clock time. What therefore are the SI units of velocity?

****

velocity is rate of change of position with respect to clock time = (change in position) / (change in clock time).

position has units of meters so change in position has units of m

clock time has units of seconds so change in clock time has unit of s

Therefore velocity has units of m / s.

#$&*

Acceleration is the rate of change of velocity with respect to clock time. What therefore are the SI units of acceleration?

****

acceleration is rate of change of velocity with respect to clock time = (change in vel) / (change in clock time).

change in vel has units of m/s

change in clock time has unit of s

so acceleration has units of (m/s)/s = m/s^2.

#$&*

A Newton is the SI unit of force. Force = mass * acceleration. In terms of kg, meters and seconds, what therefore are the units of of force? The resulting units are the units of a Newton.

****

mass is in kg, acceleration in m/s^2, so force is in kg * m/s^2

#$&*

A Joule is the SI unit of work. Work = force * displacement. In terms of kg, meters and seconds, what therefore are the units of work? The resulting units are the units of a Joule.

****

Force is measured in kg m/s^2, displacement in meters, so work is in kg m/s^2 * m = kg m^2 / s^2.

#$&*

Power is measured in watts. A watt is a Joule per second. What therefore are the fundamental units equivalent to a watt?

****

work or energy is measured in kg m^2 / s^2 (i.e., Joules), time in s, so power is in units of Joules / sec = (kg m^2 / s^2) / s = kg m^2 / s^3.

#$&*

What are the fundamental units equivalent to Newtons * meters? What quantity would be expressed in these units?

****

A Newton is a kg m/s^2, so N * m is kg m/s^2 * m = kg m^2 / s^2.

This was seen earlier to be the fundamental units of the Joule.

#$&*

What are the units of kinetic energy?

****

ke=1/2kg*v^2

KE = 1/2 m v^2

m is in kg, v in m/s

So KE is in kg * (m/s)^2 = kg * m^2 / s^2.

Note that this was earlier shown to be equivalent to the Joule.

#$&*

`q002. Sketch an object on inclines of 10, 20 and 30 degrees; you will have three sketches. For each incline, sketch the arrow indicating the force exerted on the object by gravity (for each incline we'll refer to this as the 'first arrow'). For each sketch, construct the arrow which represents the component of the gravitational force parallel to the incline (you do this by projecting the arrow at a right angle onto a line parallel to the incline, as we did in class).

For each incline, estimate the length of this arrow as a percent of the first. Give your three estimated percents:

****

An accurate sketch would show that the percents are about 17%, 34% and 50%.

Discrepancies could be due to errors in estimating the angle of the incline, projecting at an angle other than perpendicular to the line of the incline, or simply errors in estimating one vector as a percent of the other.

#$&*

`q003. My truck has a mass of about 1400 kg. It reaches the part of the road which has a nearly constant incline, in front of VHCC, at a speed of 5 meters / second in the direction down the incline. In 12 seconds its speed has increased to 10 meters / second.

What is my acceleration on that incline?

****

acceleration is ave roc of vel wrt clock time = change in vel / change in clock time = (vf - v0) / `dt = (10m/s-5m/s)/12s=5m/s)/12s=.417m/s^2

#$&*

What is the net force on the truck as it coasts down that incline?

****

F_net = m a = 1400kg*.417m/s^2=583.33kg*m/s^2.

Note that kg m/s^2 is the fundamental unit of the Newton, so the net force is about 580 N.

#$&*

What is the change in its kinetic energy as it coasts down the incline?

****

KE_0 = 1/2 m v0^2 = 1/2 1400*(5m/s)^2= 700kg*25m^2/s^2=17500kg m^2/s^2

KE_f = 1/2 m vf^2 = 1/2 1400*(10m/s)^2= 700kg*100m^2/s^2=70000kg m^2/s^2

`dKE = KE_f - KE_0 = 70000kg m^2/s^2-17500kg m^2/s^2=52500kg m^2/s^2

#$&*

If friction exerts a force opposite to my direction of motion, with the magnitude of the frictional force equal to 2% of the force exerted by gravity on the truck, then during this interval how much work is done on the truck by the frictional force?

****

The acceleration of gravity is 9.8 m/s^2 so the force exerted by gravity on the truck is

F_grav = m a_grav = 1400 kg * 9.8 m/s^2 = 13 700 kg m/s^2.

Note that this is the same as 13 700 Newtons.

Now, 2% of this gravitational force is about 270 Newtons.

The work by the frictional force can be denoted `dW_frict. Using this notation we conclude that `dW_frict = F_frict * `ds.

We need to find `ds for this motion.

The average velocity is (5 m/s + 10 m/s) / 2 = 7.5 m/s, and the interval lasts 12 seconds, so the truck moves 7.5 m/s * 12 s = 90 meters.

So F_frict * `ds = 270 N * 90 m = 2500 N * m = 2500 kg m^2 / s^2.

This is actually not quite right. The frictional force is in the direction opposite the displacement. The signs of F_frict and `ds are therefore opposite, and we conclude that the work done by friction is

`dW_frict = -2500 kg m^2 / s^2, or -2500 Joules.

#$&*

The net force on the truck is the combination of the frictional force, and the component of the gravitational force which acts in the direction down the incline. What therefore is that component of the gravitational force, and what percent is this of the total gravitational force pulling the truck toward the center of the Earth?

****

The net force on the truck was found earlier to be about 580 Newtons; since the truck is speeding up this net force is in the direction of motion.

The frictional force was just found to be about 270 Newtons, and is in the direction opposite motion.

The net force is the sum of the frictional force and the 'parallel component' of the gravitational force, so

F_net = F_parallel + F_frict.

It follows that the parallel component of the gravitational force is F_net - F_frict.

Choosing the direction of motion as positive, we have

F_parallel = 580 Newtons - (-280 Newtons) = 580 N + 280 N = 860 N.

In fundamental units this is 860 kg m^2 / s^2.

#$&*

Make a reasonably accurate sketch depicting the incline, the truck, the gravitational force and its component along the incline.

****

#$&*

`q004. In the preceding, suppose that the direction down the incline is positive. The 5 meter / second initial velocity is therefore positive, and we would write v_0 = + 5 m/s. If the force of air resistance on the car was 2 kg m/s^2, then since that force is directed up the incline, it would be represented as -2 kg m/s^2.

Give each of the following, including its sign, its numerical value and its units:

The final velocity of the truck.

****

The final velocity is in the direction down the incline so is positive.

Thus vf = +10 m/s.

#$&*

The change in the kinetic energy of the truck.

****

As seen before, the change in KE is KE_f - KE_0. This was shown earlier to be about 50 000 kg m^2 / s^2, or 50 000 Joules.

#$&*

The acceleration of the truck.

****

The acceleration is in the direction of motion, since the truck speeds up.

Thus, as found earlier, a = +.417m/s^2

#$&*

The force of friction on the truck.

****

The force of friction was found earlier to have magnitude about 280 N.

This force is in the direction opposite motion.

So the frictional force is -280 N.

#$&*

The component of the gravitational force parallel to the direction of the incline.

****

This was found earlier to be about 860 Newtons.

This force has to be greater than the net force, since friction contributes its negative share to the net force.

#$&*

The net force on the truck.

****

F_net = m a = 1400 kg * (+.417 m/s^2) = +580 N, or +580 kg m/s^2.

#$&*

`q005. Suppose the truck coasts up the incline, starting at a velocity of 15 m/s, and continues until its velocity has decreased to 10 m/s. The frictional force still has a magnitude equal to 2% of the total gravitational force on the truck.

Let the direction down the incline be positive.

Give each of the following, including its sign, its numerical value and its units:

The final velocity of the truck.

****

vf is still + 10 m/s.

#$&*

The change in the kinetic energy of the truck.

****

KE_0 = 1/2 m v0^2 = 1/2 1400*(10m/s)^2=700kg*100m^2/s^2=70000kg*m^2/s^2

KE_f = 1/2 m vf^2 = 1/2 1400*(15m/s)^2=700kg*225m^2/s^2=157500kg*m^2/s^2

So

`dKE = KE_f - KE_0 = 70000kg*m^2/s^2-157500kg*m^2/s^2=-87500kg*m^2/s^2

#$&*

The force of friction on the truck.

****

The force of friction is, as before, about -280 N.

#$&*

The component of the gravitational force parallel to the direction of the incline.

****

This is as before about +860 N, or +860 kg m/s^2.

#$&*

The net force on the truck.

****

The net force on the truck is unchanged, provided we continue to neglect air resistance.

#$&*

The acceleration of the truck.

****

The net force on the truck is unchanged so its acceleration is unchanged. Still + .417 m/s^2.

#$&*

`q006. This problem requires that you use the four equations of uniformly accelerated motion. If the truck reaches an incline on which its acceleration is .5 m/s^2, with velocity 10 m/s (both velocity and acceleration in the same direction), then how long will it take to reach a point 200 meters down the incline and how fast will it be moving at that point? You will need to carefully identify which of the quantities v0, vf, ds, dt and a are given. Then you should jot down the four equations of uniformly accelerated motion and select the one that most easily gives you additional information, and proceed from that point.

****

.5 m/s^2 is the acceleration a.

10 m/s is its initial velocity v0.

200 meters is its displacement `ds.

If downward is the positive direction then all these quantities are positive.

Given a, v0 and `ds we can use the third or fourth equation to obtain additional information.

Using the fourth equation

vf^2 = v0^2 + 2 a `ds

we find that

vf = +- sqrt( v0^2 + 2 a `ds) = +- sqrt( (10 m/s)^2 + 2 * .5 m/s^2 * 200 m) = +- sqrt( 300 m^2 / s^2) = +- 17.3 m/s.

This velocity is downward, so we discard the negative solution and conclude that

vf = 17.3 m/s.

Its average velocity is therefore

vAve = (10 m/s + 17.3 m/s) / 2 = 13.7 m/s.

It travels the 200 meters in time interval `dt, such that vAve = `ds / `dt. Thus

`dt = `ds / vAve = 200 m / (13.7 m/s) = 15 sec, very roughly.

#$&*

`q007. This problem is fairly challenging. I expect all University Physics students to get it, and hope that at least some General College Physics students will also get it despite the fact that it's probably at least a little bit too challenging for this point of the course.

Suppose the truck is moving up the incline, on which its acceleration is .7 m/s^2 down the incline. It passes you moving at 12 meters / second, and then passes your friend, who is standing 16 meters up the incline from you.

How long does it take the truck to travel the intervening distance?

****

We can choose either up or down the incline as positive.

Let's choose up as positive.

Then the acceleration, being down the incline, is

-.7 m/s^2.

The car is initially moving up the incline so its initial velocity is

v0 = +12 m/s.

The displacement is also up the incline so

`ds = + 16 m.

Writing down the four equations of motion we find that we can most easily get additional information using the fourth equation

vf^2 = v0^2 + 2 a `ds

which yields

vf = +- sqrt( v0^2 + 2 a `ds) = +- sqrt( (12 m/s)^2 + 2 * (-.7 m/s) * 16 m) = +- sqrt(144 m^2 / s^2 - 22 m^2 / s^2) = +- sqrt(122 m^2 / s^2) = +-11 m/s, approximately.

The truck is still moving up the hill so

vf = + 11 m/s.

Its average velocity on the interval is

vAve = (vf + v0) / 2 = (12 m/s + 11 m/s) / 2 = 11.5 m/s, so the 16 meter displacement requires time

`dt = `ds / vAve = 16 m / (11.5 m/s) = 1.4 s, approx.

#$&*

How far up the incline does it go before coming to rest?

****

For the interval on which it comes to rest, starting at your position, we have vf = 0.

v0 is still 12 m/s and a is still -.7 m/s^2.

We can easily reason this out:

vAve = (vf + v0) / 2 = (12 m/s + 11.5 m/s) / 2 = 11.75 m/s.

`dv = vf - v0 = 0 - 12 m/s = - 12 m/s.

Since a = `dv / `dt, we have `dt = `dv / a = -12 m/s / (-.7 m/s^2) = 17 sec, approximately.

At average velocity 11.75 m/s, in 17 s the car travels about

`ds = vAve * `dt = 11.75 m/s * 17 s = 190 m.

#$&*

If it then coasts back down the incline, accelerating at .5 m/s^2, how long will it take to travel from the position of your friend to your position?

****

The car has 190 meters to travel back to you, starting from rest.

For the corresponding interval, still using up as positive, we have

`ds = -190 m

a = -.5 m/s^2

v0 = 0

so that

vf = +- sqrt( v0^2 - 2 a `ds) = ... = +- sqrt(190 m^2 / s^2) = +- 13.7 m/s, very approximately.

The car reaches you with a downward velocity, so

vf = - 13.7 m/s.

To analyze the motion between your friend's position and yours we set up a new interval. The initial event is the car reaching your friend, the final event is the car reaching you.

For this interval, still using upward as positive

vf = -13.7 m/s

a = -.5 m/s^2

`ds = -16 meters.

The fourth equation again works:

vf^2 = v0^2 + 2 a `ds

Solving for v0 we get

v0^2 = vf^2 - 2 a `ds

so that

v0 = +- sqrt(vf^2 - 2 a `ds ) = +- sqrt((13.7 m/2)^2 - 2 * (-.5 m/s^2) * (-16 m) ) = +- sqrt( 174 m^2 / s^2) = +- 13.3 m/s, approx..

This velocity is downward, so it's negative and

v0 = -13.3 m/s.

The car reaches your friend at 13.3 m/s in the downward direction, and you at 13.7 m/s in the same direction, therefore averaging 13.5 m/s for a displacement of 16 meters (both down the incline).

The time required is therefore

`dt = `ds / vAve = 16 m / (11.5 m/s) = 1.4 sec, approx.

However if calculated more accurately, there is a difference between the time required for the car to travel up the incline past you and your friend, and the time required to travel the same distance coming down.

*@

@& You're doing well. However do check your work against the appended document, just above this note.

You do need to use the equations on some of the later questions.*@