1006

General College Physics 6th Edition Asst:

Chapter 2 problems 8, 11, 14, 16, 24, 28, 39, 50, 54, 55

University Physics:  Text Problems as Assigned in Assts 10-11 on Homepage.

At h:\shares\physics run the Multiple Choice Questions choosing the 201_core 'course'.  Run the following:

• Core Questions to Date

Experiment:

Using the same setup as last time take the measurements you need to determine

• the work required against gravity to raise the ball from the bottom of the ramp to the top of the ramp
• the KE of the ball at the end of the ramp
• the percent of the work required to raise the ball represented by the ball's KE at the end of the ramp

From your results and the length of the ramp determine the average net force on the marble on each ramp.

Graph average net force vs. ramp slope.

Link to Spreadsheet for experiment results obtained in class.  The spreadsheet is in a tab-delimited text file and should easily open in your spreadsheet.

Quiz Results:

Approximately 75% of the quiz questions were answered correctly.

The table below shows questions for which the wrong answer was given by students on today's quiz.  If an answer was given by more than one student then it will appear more than once.

You should be sure you understand the error that led to the incorrect answer.

Correct Answer	Question
y	If `ds = (vf + v0) / 2 * `dt an equation of uniformly accelerated motion?
y	If `ds = (vf + v0) / 2 * `dt an equation of uniformly accelerated motion?
n	Is `ds = (vf - v0) / 2 * `dt  an equation of uniformly accelerated motion?
n	The average rate of change of acceleration with respect to clock time is average position.
y	Is vf = v0 + a `dt an equation of uniformly accelerated motion?
y	Is vf = v0 + a `dt an equation of uniformly accelerated motion?
y	Average acceleration * change in clock time = change in velocity.
y	Average acceleration * change in clock time = change in velocity.
n	Is `ds = v0 `dt + 2 a `ds an equation of uniformly accelerated motion?
y	Is `ds = v0 `dt + .5 a `dt ^2 an equation of uniformly accelerated motion?
n	Is `ds = v0 `dt + .5 a `dt an equation of uniformly accelerated motion?
n	Is `ds = v0 `dt + .5 a `dt an equation of uniformly accelerated motion?
n	Is `ds = v0 `dt + .5 a `dt an equation of uniformly accelerated motion?
n	Is vf^2 = v0^2 + .5 a `dt ^2 an equation of uniformly accelerated motion?
n	Is vf^2 = v0^2 + .5 a `dt ^2 an equation of uniformly accelerated motion?
y	Is vf^2 = v0^2 + 2 a `ds an equation of uniformly accelerated motion?
y	Is vf^2 = v0^2 + 2 a `ds an equation of uniformly accelerated motion?
n	Average acceleration = average velocity / change in clock time.
n	Change in position = change in velocity * change in clock time.
y	Final acceleration = initial acceleration + change in acceleration
n	average acceleration = initial acceleration + change in acceleration
y	aAve = `dv / `dt .
y	aAve = `dv / `dt .
y	vAve = `ds / `dt .
y	vAve * `dt = `ds
y	vAve * `dt = `ds
n	aAve * `dt = `ds .
n	aAve = vAve / `dt .
y	vf = v0 + `dv .
n	Is `ds = (vf - v0) / 2 * `dt  an equation of uniformly accelerated motion?
n	Is `ds = (vf - v0) / 2 * `dt  an equation of uniformly accelerated motion?
n	Is `ds = (vf - v0) / 2 * `dt  an equation of uniformly accelerated motion?
n	`Is ds = (vf + v0) / `dt * 2  an equation of uniformly accelerated motion?
y	Is vf = v0 + a `dt an equation of uniformly accelerated motion?
y	Is vf = v0 + a `dt an equation of uniformly accelerated motion?
y	Average velocity * change in clock time = change in position.
n	Is `ds = v0 `dt + 2 a `ds an equation of uniformly accelerated motion?
n	Is `ds = v0 `dt + .5 a `dt an equation of uniformly accelerated motion?
n	Is vf^2 = v0^2 + .5 a `dt ^2 an equation of uniformly accelerated motion?
n	Is vf^2 = v0^2 + .5 a `dt ^2 an equation of uniformly accelerated motion?
y	Is vf^2 = v0^2 + 2 a `ds an equation of uniformly accelerated motion?
y	Is vf^2 = v0^2 + 2 a `ds an equation of uniformly accelerated motion?
n	Average acceleration = average velocity / change in clock time.
y	aAve = `dv / `dt .
y	vAve = `ds / `dt .
n	aAve = `ds /`dt .
y	vf = v0 + `dv .
y	vf = v0 + `dv .

My velocity at the beginning of the uniform-acceleration event was somewhere between 10 m/s and 15 m/s.  My acceleration was -5 m/s^2 and the uniform-acceleration event terminated after 8 meters.

What is the possible range of final velocities?

Given v0, a, `ds we use vf^2 = v0^2 + 2 a `ds to get the final velocity vf = +- sqrt( v0^2 + 2 a `ds).

For the faster initial velocity we know that v0 = 15 m/s, a = -5 m/s^2 and `ds = 8 m so we get vf = 12.2 m/s.

For v0 = 10 m/s, same a and `ds we get vf = 4.4 m/s.

What is the range of KE changes, assuming vehicle mass 1300 kg?

The initial KE in the first case was .5 m v0^2 = .5 * 1300 kg * (15 m/s)^2 = 145,000 kg m^2 / s^2 = 145,000 Joules, approx..

Final KE at 12.2 m/s (first case) is .5 m v0^2 = .5 * 1300 kg * (15 m/s)^2 = 94225 kg m^2 / s^2 = 94,000 Joules.

Change in KE in the first case is about 94,000 J - 145,000 J = -51,000 J.

The initial KE in the first case was .5 m v0^2 = .5 * 1300 kg * (10 m/s)^2 = 65,000 kg m^2 / s^2 = 65,000 Joules, approx..

Final KE at 4.5 m/s (second case) is .5 m v0^2 = .5 * 1300 kg * (4.5 m/s)^2 = 13,000 kg m^2 / s^2 = 13,000 Joules.

In the second case the change in KE is about 13,000 J - 65,000 J = -52,000 J.

The changes in KE are within mental error the same.

What is the range of final kinetic energies?

The range of final KE is 13,000 J to 94,000 J, approx..

How much work was done by the net force acting on the vehicle?

The net force is Fnet = m a = 1300 kg * (-5 m/s^2) = -6500 N.

The work done over 8 meters by this net force is

`dWnet = -6500 N * (8 m) = -52,000 J.

This is identical to the change in KE in both cases.

Experiment

To fall to the floor from a height of 94 cm, provided initial vertical velocity is zero, takes the same time whether the object starts with 0 horizontal velocity or with a nonzero horizontal velocity.

How long does it take to fall 94 cm from table edge to floor?

v0 = 0, a = 9.8 m/s^2 and `ds = .94 m.  Using the 4th equation to find vf then the first equation to find `dt we get `dt = .43 second (vf was about 4.3 m/s).

The ball traveled 46 cm in the horizontal direction as it fell to the floor.

The only force acting in the horizontal direction is air resistance, which at this speed is small, and for the mass of the ball results in negligible acceleration in the horizontal direction.

What therefore was the velocity of the ball as it left the edge of the table?

We know horizontal displacement is 46 cm and time interval is .94 sec.  We immediately calculate average horizontal velocity

• vAve = `ds / `dt = 46 cm / (.43 s) = 106 cm/s or 1.06 m/s.

Assuming the ball has mass 100 grams by how much does its KE change as it rolls from rest down the ramp?

The initial KE is zero.

The final KE is .5 m v^2 = .5 * (.1 kg) ( 1.06 m/s)^2 = .056 Joules.

How much work do I have to do against gravity to lift the ball from the tabletop to the starting point on the ramp, a vertical distance of 10 cm?

The force I have to exert is equal and opposite to the force of gravity, so the force is F = - Fgrav.  The work I have to do against gravity is therefore `dWagainstGrav = -Fgrav * `ds.

Fgrav = -m g = -.100 kg * 9.8 m/s^2 = -.98 N.

So

`dWagainstGrav = - ( -.98 N) * .1 m = .098 Joules.

We see that the KE change of the ball is .056 Joules and the work required to lift the ball is .098 Joules.  Is this consistent with the idea of energy conservation?

Do the following: 

• Set up a similar system.
• Take the measurements you will need to repeat the above analysis (altitude change on ramp, distance of fall, horizontal displacement during fall).
• Report the work required to lift the ball (assume a 100 g mass, whatever ball you use) and the KE change.
• Report the KE change as a percent of the work required to raise the ball.