Question

A block initially at rest is allowed to slide down a frictionless ramp and attains a speed v at the bottom. To achieve a speed 2v at the bottom, how many times as high must a new ramp be?

Answer 1

Answer:

new height of ramp must be 4 times that of the first ramp.

Explanation:

From conservation of energy, we know that;

Potential energy at the top of ramp = kinetic energy at the bottom of ramp.

Thus;

mgh_t = ½mv²

m will cancel out to give;

gh_t = ½v²

Thus means that the height of the first ramp is directly proportional to the square of the speed.

Thus;

h_t ∝ v²

Now, for the new ramp, we are told that we want to achieve a speed of 2v at the bottom.

Thus;

h'_t ∝ (2v)²

h'_t ∝ 4v²

From earlier we saw that;

h_t ∝ v²

Thus;

New height of ramp is;

h'_t ∝ 4h_t

Thus, new height of ramp must be 4 times that of the first ramp.

Answer 2

To increase the speed to 2v at the bottom of the ramp, the height will be 4 times higher.

The given parameters;

• velocity at the bottom ramp, = v
• let the height of the ramp = h

Apply the principle of conservation of energy;

P.E = K.E

$mgh = \frac{1}{2} mv^2\\\\gh = \frac{1}{2} v^2\\\\h = \frac{v^2}{2g}$

To increase the speed 2v at the bottom of the ramp, the height will be;

$H = \frac{(2v)^2}{2g} \\\\H = \frac{4v^2}{2g} \\\\H = 4(\frac{v^2}{2g} )\\\\H = 4(h)$

Thus, to increase the speed to 2v at the bottom of the ramp, the height will be 4 times higher.

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