# Physics block speeds problem

#### 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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#### HallsofIvy

Homework Helper
Here is what I did so far

KE =1/2mv^2
M= 1.0 kg
V= ?
Wsubg= mgh
Since the object is at rest initially, V= 0 and the potential energy is 1(9.8)(9)= 88.2 Joules

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

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
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.

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
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.

#### Physicsit

Clarification

One quick question before I get started

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

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

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