You stand at the bottom of an 8.0-m-long ramp that is inclined at 37∘ above the horizontal. You grab packages off a conveyor belt and propel them up the ramp. The coefficient of kinetic friction between the packages and the ramp is μk=0.30.

Question

A)
What speed do you need to give a package at the bottom of the ramp so that it has zero speed at the top of the ramp?
Express your answer using two significant figures.

B)
Your coworker is supposed to grab the packages as they arrive at the top of the ramp. But she misses one and it slides back down. What is its speed when it returns to you?
Express your answer using two significant figures.

___________________________________

I have broken the Fg into vertical and horizontal components.
Vertical: Wcos(37)
Horizontal: Wsin(37)
Friction is therefore 0.3*Wcos(37)

I don't quite know what to do after this, maybe using energy equations?
mgh = 0.5mv^2, where v = radical 2gh?
But there is an angle involved, and a distance.

For B, the returning velocity would be the initial velocity minus the speed lost to friction going up and then down. I do not know how to calculate that either.

Any help is appreciated, thanks.

Answer 1

you could do it energy.
initial energy:1/2 m v^2
that initial energy is converted to Friction and PE.
friction:.3*mg*cos37*8
PE: mg*8*sin37

set initial energy=friction+PE

second part.
Initial PE=mg*8*sin37
friction=.3mg*cos37
final KE=1/2 m v^2
finalKE=Initial PE-friction
solve for v

Answer 2

For the second part, I set:

1/2mv^2 = (mg*8*sin37) - (0.3*mg*cos37)
or
0.5v^2 = (g*8*sin37) - (0.3g*cos37)
0.5v^2 = (47.18) - (2.347) = 44.833
v^2 = 2(44.833) = 89.666
v = rad(89.666) = 9.469 m/s

but this is not the correct answer. Did I miss something?

Answer 3

You forgot distance on the friction component. Energy= friction*distance

Answer 4

Ah, so I did.
Thanks for the help!