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?
Answer:
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