A 2.5-kg block slides down a 17.0° inclined plane with constant acceleration. The block starts from rest at the top. At the bottom its velocity reaches 0.70 m/s. The length of the incline is 1.7 m.

Question

A 2.5-kg block slides down a 17.0° inclined plane with constant acceleration. The block starts from rest at the top. At the bottom its velocity reaches 0.70 m/s. The length of the incline is 1.7 m.
What is the acceleration of the block?
What is the coefficient of friction between the plane and the block?

Henry · Answer

Given:
M = 2.5kg. = Mass of block.
A = 17o.
Vo = 0 = Initial velocity.
V = 0.70 m/s = Final velocity.
d = 1.7 m. = Length of incline.

M*g = 2.5 * 9.8 = 24.5 N. = Wt. of block.

Fp = 24.5*sin17 = 7.16 N. = Force parallel with incline.

Fn = 24.5*cos17 = 23.43 N. = Normal force.

Fk = u*Fn = u*23.43 = 23.43u.

a. V^2 = Vo^2 + 2a*d.
0.7^2 = 0 + 2a*1.7,
a = 0.144 m/s^2.

b. Fp-Fk = M*a.
7.16-23.43u = 2.5*0.144.
u = ?