CQ_1_141

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The tension force is in the direction opposite this displacement, so the force does negative work.

At this point how fast will the domino be moving?

answer/question/discussion: v = sqroot(2KE/m) = (2*.06 / .02) = sqroot(6) =2.4m/s

If KE was zero then v would have to be zero.

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Good approach, with a few errors in detail. Ignore the request for a resubmission, unless you have questions:

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A rubber band begins exerting

• Between the 8 cm and 10 cm length, what are

  the minimum and maximum tensions, and what do you think is the average tension? 

The tension increases from 0 N at the 8 cm

length to 3 N at the 10 cm length. 

If tension is a linear function of length,

then the average tension is (0 N + 3 N) / 2 = 1.5 N.

It is unlikely that a rubber band has a

perfectly linear tension function, but without further information this is

the most reasonable approximation to make, and it probably isn't much in

error.

• How much work is required

  to stretch the rubber band from 8 cm to 10 cm? 

The total displacement is 10 cm - 8 cm = 2 cm.

The force changes linearly with displacement, from 0 N to 3 N, so the average

force is (0 N + 3 N) / 2 = 1.5 N.

The work is done by the stretching force is therefore

• work exerted on rubber band = force * displacement = 1.5 N * 2 cm =

  3.0 N * cm, or .03 Joules.

• During the stretching process is the tension force in the

  direction of motion or opposite to the direction of motion? 

• Does the tension

  force therefore do positive or negative work?

The stretching force is in the direction of

motion and does positive work. 

However the tension force is the force

exerted by, not on, the rubber band, and is in the direction opposite

motion.  The tension force therefore does negative work during the

stretching process.

The rubber band is released and as

• Again assuming that the

  tension force is conservative, how much work does the tension force
  do on the domino? 

If the force is conservative, then the work

done on return is equal and opposite to the work done in the stretching

process.  That work was negative, so the work done on return is

positive. 

We conclude that the work done by the

tension force is on return is +.03 Joules

• Assuming this is the only force acting on the domino, what

  will then be its kinetic energy when the rubber band reaches its 8 cm length? 

Given the stated assumption, the tension

force is the net force. 

The work done by the net force acting on an

object or system is equal to the change in KE of that object or system: 

• `dW_net = `dKE 

Since `dW_net is in this case +.03 Joules,

`dKE = +.03 J and the domino's KE increases by .03 J. 

The domino was originally at rest, so its

KE increases from 0 to .03 Joules.

• At this point how fast will the domino be moving?

KE = 1/2 m v^2, so v = sqrt(2 * KE / m). 

Thus when its KE is .03 Joules, the .02 kg domino is moving at velocity

v = sqrt( 2 * .03 Joules / (.02 kg) ) =

sqrt(3 m^2 / s^2) = 1.7 m/s, approx.