Question

Using conservation of energy, find the speed vb of the block at the bottom of the ramp.

Express your answer in terms of some or all the variables m, v, and h and any appropriate constants.

Answer 1

For the first problem, I'll assume the surfaces are frictionless.

The total energy E of the block at any point is:

E = U + K

where U is the gravitational potential energy and K is the kinetic energy.

We also have:

U = mgh

where m is the mass, g is the gravitational acceleration and h is the height.

And

K = (1/2)mv^2

is the kinetic energy.

When the block is at the top of the ramp its total energy is therefore

E = mgH + (1/2)mv^2

where H is the height of the ramp and v is its speed at the top.

At the bottom of the ramp, the energy is:

E = 0 + (1/2)m v_f^2

where v_f is the final speed, and there is no potential energy.

Energy is conserved in this example, since the system of Earth+block+ramps is isolated, so the E's are constant in both cases. Therefore:

mgH + (1/2)m v^2 = (1/2) m v_f^2

Re-arranging to find v_f, we get:

2gH + v^2 = v_f^2

or

v_f = sqrt(v^2 + 2gH)