Query Questions 17

#$&*

course Phy 241

October 15 around 5:30pm

017. `query 17

*********************************************

Question: `qquery Univ. 7.42 (7.38 in 10th edition). 2 kg block, 400 N/m spring, .220 m compression. Along surface then up 37 deg incline all frictionless. How fast on level, how far up incline?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

400(0.220)=88N at 0.220m mark

Favg=88/2=44N

‘dW=Favg*’ds

=44N*0.220m

=9.68J

V=sqrt (2KE/m)

=sqrt (2(9.68)/2)

=3.11m/s is how fast

Sin 37 degrees=0.602

‘ds=KE/0.602*(ma)

‘ds=(9.68)/11.808

‘ds=0.82meters is how far up the incline it went

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

`a** The spring exerts a force of 400 N / m * .220 m = 88 N at the .220 m compression. The average force exerted by the spring between equilibrium and this point is therefore (0 N + 88 N) / 2 = 44 N, so the work done in the compression is

`dW = Fave * `ds = 44 N * .220 m = 9.7 Joules, approx.

If all this energy is transferred to the block, starting from rest, the block's KE will therefore be 5.0 Joules. Solving KE = .5 m v^2 for v we obtain v = sqrt(2 KE / m) = sqrt(2 * 9.7 Joules / (2 kg) ) = 3.2 m/s, approx..

No energy is lost to friction so the block will maintain this speed along the level surface. As it begins to climb the incline it will gain gravitational PE at the expense of KE until the PE is 9.7 J and the KE is zero, at which point it will begin to slide back down the incline.

After traveling through displacement `ds along the incline the height of the mass will be `ds sin(37 deg) = .6 `ds, approx., and its gravitational PE will be PE = m g h = m g * .6 `ds = .6 m g `ds.

Setting this expression equal to KE we obtain the equation

.6 m g `ds = KE,

which we solve for `ds to obtain

`ds = KE / (.6 m g) = 9.7 Joules / (.6 * 2 kg * 9.8 m/s^2) = .82 meters, approx. **

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

OK

------------------------------------------------

self-critique rating #$&*:

*********************************************

Question: `qquery univ phy 7.50 62 kg skier, from rest, 65 m high. Frict does -10.5 kJ.

What is the skier's speed at the bottom of the slope?

After moving horizontally over 82 m patch, air res 160 N, coeff frict .2, how fast is she going?

Penetrating 2.5 m into the snowdrift, to a stop, what is the ave force exerted on her by the snowdrift?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Skier’s speed at bottom of slope:

62(9.807)(65)=39522.21J or 39.5kJ

KE=39.5kJ-10.5kJ=29kJ

V=sqrt (2KE/m)

=sqrt (2(29000)/62)

=30.6m/s

How fast?

At 82m: 0.2(62)(9.807)=120N

160N+120N=280N

280N*82m=22960J or 22.9kJ

KE of skier: 29kJ-22.9kJ=6kJ

Speed: v=sqrt (2(6000)/62)

=13m/s

Average force exerted on her:

‘dW=Favg*’ds

Favg=’dW/’ds

=(6000J)/2.5m

=2400N

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

`a** The gravitational PE of the skier decreases by 60 kg * 9.8 m/s^2 * 65 m = 38 kJ, approx. (this means 38 kiloJoules, or 38,000 Joules). The PE loss partially dissipated against friction, with the rest converted to KE, resulting in KE = 38 kJ / 10.5 kJ = 27.5 kJ.

Formally we have

`dKE + `dPE + `dWnoncons = 0, where `dWnoncons is the work done by the skier against friction. Since friction does -10.5 kJ of work on the skier, the skier does 10.5 kJ of work against friction and we have `dKE = -`dPE - `dWnoncons = - (-38 kJ) - 10.5 kJ = 27.5 kJ.

The speed of the skier at this point will be

v = sqrt( 2 KE / m) = sqrt( 2 * 27,500 J / (65 kg) ) = 30 m/s, approx.

Over the 82 m patch the force exerted against friction will be .2 * 60 kg * 9.8 m/s^2 = 118 N, approx., so the force exerted against nonconservative forces will be 118 N + 160 N = 280 N approx.. The work done will therefore be

`dWnoncons = 280 N * 82 m = 23 kJ, approx.,

and the skier's KE will be

KE = 27.5 kJ - 23 kJ = 4.5 kJ, approx.

This implies a speed of

v = sqrt( 2 KE / m) = 12 m/s, approx.

To stop from this speed in 2.5 m requires that the remaining 4.5 kJ of KE be dissipated in the 2.5 m distance. Thus we have

`dW = Fave * `ds, so that

Fave = `dW / `ds = 4500 J / (2.5 m) = 1800 N.

This is a significant force, about 3 times the weight of the skier, but distributed over a large area of her body will cause a good jolt, but will not be likely to cause injury.**

STUDENT QUESTION

If the PE is = 20.6, then why is the initial KE= 20.6 so that we are adding the .245 to the initial KE of 20.6 to get

20.8, I thought that the KE was equal and opposite to the PE why would we not subtract here?

INSTRUCTOR RESPONSE

`dKE + `dPE = 0, provided there are no nonconservative forces acting on the system.

In such a case, PE goes up as the mass rises, so KE goes down. Another way of looking at it: All or part of the KE converts to PE.

The mass can only go as high as the initial KE permits. Once the initial KE is 'used up', no increase in PE is possible (recall the assumption that no nonconservative forces act during this phase of motion).

At maximum height the mass is still moving in the horizontal direction, so not all of the KE converts to PE.

In this case PE increases by 20.6 M m^2/s^2, .245 M m^2/s^2 of KE is still present at the highest point, so about 20.8 M m^2 / s^2 of KE must have been present initially.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

OK

It looks like I did everything right. Our final answer came out a lot different, so it could just be rounding errors and sig figs. I also think you did 60kg for your skier and I used 62kg?!

60 kg vs. 62 kg makes a little difference, but not much

A greater difference is approximation errors in my calculation of initial speed and KE. Since most of the KE is dissipated by the time she hits the drift, a relatively small error in the (large) PE loss becomes a relatively larger error in the (smaller) KE remaining to be dissipated.

In any case your reasoning is all correct, and I trust your numbers more than mine.

------------------------------------------------

self-critique rating #$&*:

Excellent. See my notes.

#$&*