Friction force on a box and roller coaster speed at bottom of hill

[shin777]

1. What force is needed to pull a 10kg box across a floor that has a coefficient of friction of .34?

my answer
F = W * f * x
F = .34(10)(9.8) = 33N

2. The carts of a roller coaster are traveling 5 m/s over the top of the first hill. If there is no energy loss due to friction and the hill is 50m high, what speed will the carts have at the bottom of the hill?

my answer

(1/2)(Vi^2 + gh) = (1/2)(Vf^2)
(1/2)[(5^2)+9.8(50)] = (1/2)(Vf^2)
Vf = 31 m/s

do they look ok? if I am wrong please correct me. :(

 

[haruspex]

First part is OK.

(1/2)(Vi^2 + gh) = (1/2)(Vf^2)

The equation is not correct. What is the expression for potential energy?

(1/2)[(5^2)+9.8(50)] = (1/2)(Vf^2)
Vf = 31 m/s

That answer does not follow from the equations you give, but it is close to the correct answer. Maybe you made a mistake in typing the equations in your post.

 

[shin777]

err.. yeah.. I think it should have been
Vi^2 + gh = (1/2)Vf^2
5^2 + 9.8(50) = (1/2)Vf^2
2(515) = Vf^2
Vf = 32 m/s

does it look ok now?

 

[haruspex]

err.. yeah.. I think it should have been
Vi^2 + gh = (1/2)Vf^2

You changed too much!

5^2 + 9.8(50) = (1/2)Vf^2
2(515) = Vf^2
Vf = 32 m/s

does it look ok now?

The answer is now correct to the accuracy quoted, but partly by luck. The 515 is too large.

 

[shin777]

my bad.. i got it now.
it's vf = ( 25 + 2(9.8)(50) ) ^ 1/2
vf = 1005^(1/2)
vf = 31.7 m/s

 

[haruspex]

my bad.. i got it now.
it's vf = ( 25 + 2(9.8)(50) ) ^ 1/2
vf = 1005^(1/2)
vf = 31.7 m/s

That's it.