13.07.2014 Views

PHYS 211 Recitation Review Problems: Solutions

PHYS 211 Recitation Review Problems: Solutions

PHYS 211 Recitation Review Problems: Solutions

SHOW MORE
SHOW LESS

You also want an ePaper? Increase the reach of your titles

YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.

Physics <strong>211</strong> Fall 2012 Midterm 2 <strong>Recitation</strong> <strong>Review</strong> <strong>Solutions</strong><br />

3) Cart<br />

For a lecture demo in front of class, a little cart (mass 0.5 kg) travels<br />

on a long, straight, level air track. Shown at right is a graph of<br />

velocity vs. time of the cart. Let's define "right" to be the positive<br />

direction.<br />

1. What is the net force (with correct sign) on the cart at t=3 s?<br />

A) 0 N<br />

m<br />

B) +3 N<br />

v<br />

6<br />

a s<br />

4<br />

m<br />

2<br />

C) -3 N<br />

s<br />

t<br />

1.5 s<br />

D) +2 N<br />

m<br />

F ma 0.5kg 4 2 2 N<br />

s<br />

E) -2 N<br />

+4 m/s<br />

+2 m/s<br />

-2 m/s<br />

+velocity<br />

time (s)<br />

1 2 3 4 5<br />

2. At t=2.5 seconds, how would you describe what the car is doing?<br />

A) It is moving to the right with steady speed.<br />

B) It is moving to the right with decreasing speed<br />

C) It is moving to the left with steady speed<br />

D) It is moving to the left with decreasing speed<br />

E) None of the above/not enough information<br />

3. At t=5 seconds, the air for the air track is suddenly turned off, so kinetic friction (alone) causes<br />

the cart to grind to a halt in another 0.5 sec. During that half-second when the cart is stopping,<br />

call the force of friction on the cart by the track f CT .<br />

What can you conclude about the frictional force f' TC on the track by the cart during this time?<br />

(assume the mass of the track itself is 5.0 kg, i.e. ten times the mass of the little cart)<br />

A) f' TC = -10 f CT<br />

B) f' TC = +10 f CT<br />

C) f' TC = -f CT /10<br />

D) f' TC = -f CT<br />

E) f' TC = +f CT<br />

Newton’s 3 rd Law: Action-reaction pairs are equal and opposite<br />

4. In the previous question, (during the half-second when the car stops due to friction) what is the<br />

coefficient of kinetic friction between the cart and the track?<br />

A) 0.8<br />

m<br />

B) 0.5<br />

v<br />

2<br />

a s<br />

4 m<br />

2<br />

C) 0.4<br />

s<br />

t<br />

0.5 s<br />

D) 0.2<br />

m<br />

4 2<br />

a s<br />

E) 0.1<br />

F ma mg<br />

m 0.4<br />

g 10<br />

2<br />

s<br />

Page 2 of 7

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!