block on ramp help?

Question

okay, i have a block on a ramp 34 degrees above the horizontal, of mass 20kg. the ramp and block have a kinetic friction coefficient of 0.44.
the block initially has a velocity of 4.6ms-1 up the ramp.

i need to work out the work done on the block as it slides up to its highest point on the ramp.
i can say that the velocity at its highest point is 0 ms-1.

i have got a diagram drawn of the block on the ramp
mg drawn straight down from the block
mg sin x drawn down the slope
mg cos x drawn at a right angle to the slope

so apart from friction, these would be all the forces i need to consider if the block was just released to slide down the slope.
BUT, it is moving up the slope with that initial velocity.

so could somebody show me how i would go about finding the net force here?
my thoughts are that i have to somehow find out how far the block reaches up the slope before it can go no further, but i dont know how to do this.

a little explanation please? thanks.

Answer 1

You've made a good start.

[EDIT -- This CAN be solved by analyzing the forces; but it turns out that there's an easier way; and in fact the final solution does not depend on the angle of the slope nor on the coefficient of friction. Read on: first the hard way, then the easy way.]

Whenever the block is sliding (regardless of whether it's going up or down), there is a force of friction acting OPPOSITE of the direction of motion.

So, if the direction of motion is UPslope (as it is in this problem), the direction of friction is DOWNslope.

The magnitude of the friction is the normal force times the coefficient of friction:

F_friction = Fn × μ

The normal force Fn is (mg)cosθ, as you've already figured out.

So, you've got two forces that are parallel to the slope:

1. (mg)sinθ (downslope) due to gravity;

2. Fn(μ) = (mg)cosθ(μ) (also downslope) due to friction.

The net forces _perpendicular_ to the slope cancel out (this must be true, because we observe there is no acceleration in the perpendicular direction). So they do not contribute to the net force.

So, the net force is:

Fnet = (mg)sinθ + (mg)cosθ(μ) [downslope]

As for finding the work done: You've correctly realized that work = force × distance, so it would be a good idea to find out just what the distance is. In cases where the acceleration is constant (as it is in this problem because (Fnet / m) is constant), you can use this kinematics equation to find distance in terms of speed and acceleration:

distance = ((v_final)² − (v_initial)²) ⁄ (2a)

Furthermore, you can plug in (Fnet / m) in place of "a".

When you work this out, you'll probably notice that "d" and "a" are opposite in sign (one's positive, the other negative). This just reflects the fact that the distance is UPslope, while the acceleration is DOWNslope (the same direction as the net force).

EDIT!!! -- If you work out all the equations above, you'll discover an amazing thing. The "Fnet" actually cancels out of the equation.

distance = ((v_final)² − (v_initial)²) ⁄ (2a)
distance = − (v_initial)² ⁄ (2a)
distance = − (v_initial)² ⁄ (2Fnet/m)
distance = − m(v_initial)² ⁄ (2Fnet)

And, to get Work:

Work = force × distance
Work = Fnet × −m(v_initial)² ⁄ (2Fnet)
Work = −m(v_initial)² ⁄ 2

This means the amount of work done to stop the block is equal to the amount of kinetic energy the block initially had. This makes sense; it's consistent with the conservation of energy.

All of which means: the angle of the slope and the coefficient of friction don't make one bit of difference to the final solution!

Answer 2

Block On A Ramp