- #1

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

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