A crate of mass M starts from rest at the top of a frictionless ramp inclined at an angle α above the horizontal. Find its speed at the bottom of the ramp, a distance d from where it started. Do this in two ways.

The speed of the crate at the bottom of the ramp is

The given parameters;

The normal force on the crate is calculated as follows;

Fₙ = mgcosα

The net horizontal force on the crate is calculated as follows;

mgsinα = ma

gsinα = a

The speed of the crate at the bottom of the ramp is calculated as follows;

Thus, the speed of the crate at the bottom of the ramp is

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Complete question is;

A crate of mass M starts from rest at the top of a frictionless ramp inclined at an angle α above the horizontal. Find its speed at the bottom of the ramp, a distance d from where it started. Do this in two ways: (a) Take the level at which the potential energy is zero to be at the bottom of the ramp with y positive upward. (b) Take the zero level for potential energy to be at the top of the ramp with y positive upward. (c) Why did the normal force not enter into your solution?

A) v2 = √(2g•d•sin α)

B) v2 = √(2g•d•sin α)

C) Because normal force is perpendicular to the displacement along the ramp and thus the work done is zero

Explanation:

Mass of the crate = M

Angle of ramp above horizontal = α

Distance between starting point to end point = d

A) Since, we have d and α, then the height of the ramp will be found using trigonometric ratios for a right angle triangle.

Thus;

h = d sin α

When the potential energy at the bottom of the ramp is zero, we have;

y1 = h and y2 = 0

Now, we know that total work done is given by the formula;

W_tot = K2 - K1 = W_other + W_grav

But since gravitational force is the only force acting on the crate in the direction of motion, then W_other = 0.

Now, W_grav is given by the formula;

W_grav = mgy1 - mgy2

So, we now have;

K2 - K1 = mgy1 - mgy2

Where K2 and K1 are final and initial kinetic energy respectively.

So,

½m(v2)² - ½m(v1)² = mgy1 - mgy2

Since v1 and y2 are zero, and y1 = h, then we have;

½m(v2)² = mgh

Cross multiply to get;

(v2)² = 2gh

We earlier established that h = d sin α

So,

(v2)² = 2g•d•sin α

v2 = √(2g•d•sin α)

B) when the potential energy is zero at the top of the tamp, we have;

y1 = 0 and y2 = -h

Like in solution A above, W_other = 0

Similarly,

½m(v2)² - ½m(v1)² = mgy1 - mgy2

We have, v1 and y1 are zero, and y2 = - h, then we have;

½m(v2)² = -mg(-h)

Cross multiply to get

(v2)² = 2gh

We earlier established that h = d sin α

So,

(v2)² = 2g•d•sin α

v2 = √(2g•d•sin α)

C) I did not enter the normal force into the solution because the normal force is perpendicular to the displacement along the ramp and thus the work done is zero

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