Block sliding down ramp with friction: finding landing distance

[ova5676]

Homework Statement

A 1.9-kg block slides down a curved, frictionless ramp. The top of the ramp is 1.5m above the ground; the bottom of the ramp is 0.25m above the ground. The block leaves the ramp moving horizontally, and lands distance D away.

A) What is distance D away?

B) Suppose the ramp is not frictionless. Find the distance D for the case in which friction on the ramp does 9.7 J of work on the block before it enters the horizontal (and still frictionless) section towards D.

For both A) and B) the time is 4.0s at distance D.

Homework Equations

Potential Energy (1) + Kinetic Energy (1) = Potential Energy (2) + Kinetic Energy (2)
mgh(1) + 1/2*m*v^2(1) = mgh(2) + 1/2*m*v^2(2)

(basically the first part is at the top of the ramp and the final is at the bottom of the ramp)

The Attempt at a Solution

Since the ramp is frictionless, the block is going at a constant speed. However, when I tried using the equation it didn't work out so there's probably another way.

 

[CompuChip]

You have posted the correct formula. However, the block is accelerated: it has no velocity at the top, but a positive velocity at the bottom. Obviously, m and g are given in the problem. Can you give me the appropriate values for h(1), v(1), h(2) and v(2) ?

 

[ova5676]

h(1) is 1.5m (reference point is the ground).

h(2) is 0.25m (reference point is the ground).

We don't know v(1) or v(2).

 

[Doc Al]

We don't know v(1) or v(2).

I would presume that the block starts from rest at the top of the ramp. So v(1) = 0. Use energy conservation to solve for v(2).