# Requesting Help

1. Oct 8, 2003

### Physicsit

Ok I have been workoing on this problem for at least a couple of hours with no luck

A 1.0-kg block is released from rest at the top of a curved fric-tionless track as shown in the figure below. (a) What are the speeds of the block at points A and B? (b) If the block goes on a level surface at point C with a coefficient of kinetic friction of 0.50, how far from point C will the block come to rest?

Attached is a graphical representation

Here is what I did so far

KE =1/2mv^2
M= 1.0 kg
V= ?
Wsubg= mgh

For point A
Final Kinetic energy = KEInitial - W subg
W sub g = mgh 1.0 X 9.8 X 2.0
Kfinal = 20

20-1/2(1.0)v^2
40= v^2
v= 6.3 m/s that is v of point a

for point b
Initial kenetic = 20

Final Kinetic energy = KEInitial - W subg
W sub g = mgh 1.0 X 9.8 X 6.0
final kentic = 60-20=40

40=1/2(1.0)v^2
80=v^2
v 8.9 m/s for point b

I then calculated the maximum static firction
using the coefficient of kinetic friction of 0.50
mg X coefficient friction

1.0X9.8X.50=4.9N

that is the maximum force that can be exerted wihtout the objectmoving

but I do not know how to claculate how far from point C will the block come to rest

I do not know if I did the first three parts correct either

I have been at this for a while now over 2 hours

I desperatley need help any advice would be much appreciated

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2. Oct 8, 2003

### HallsofIvy

Since the object is at rest initially, V= 0 and the potential energy is 1(9.8)(9)= 88.2 Joules

Why? I assume the "20" is an approximation to 19.8 which is 9.8*2. That's the potential energy not the kinetic energy.

At point A, the object's height is 2 so its potential energy is
1(9.8)(2)= 19.6 Joules. That's a decrease of 88.2- 19.6= 68.6 Joules which must have gone into kinetic energy: (1/2)(1)v2= 68.6 so v= 11.7 m/s.

You're going the right way but without friction the total energy is still the potential energy at the start: 88.2 Joules. At point B the height is 6 meters so its potential energy is 1(9.8)6= 58.8 Joules (you round off to 60). That's a change (from the INITIAL point) of 88.2- 58.8= 29.4 Joules. We must have (1/2)(1)v2= 29.4 or
v= 7.7 m/s at point B.

Finally, at C, the height is 0 so potential energy (relative to 0 height of course) is 0. The 88.2 Joules of energy is now all kinetic energy: (1/2)(1)v2= 88.2 so v= 13.3 m/s.

No, your calculation is correct but 0.5 is the coefficient of KINETIC friction, not static friction! Since friction is imposing a force of 4.9 N, we have f= ma or -4.9= (1.0)a. The acceleration is
-4.9 m/s2. At that acceleration the object slows from 13.3 m/s to 0 m/s in (v= at) 13.3/4.9= 2.7 seconds. Of course,
x= v0 t- (a/2) t2 so x= (13.3)(2.7)- (4.9/2)(2.7)2= 18 meters.

3. Oct 8, 2003

### Physicsit

Clarification

One quick question before I get started

why is it times 9 and not times 10 if initial height is 10?