Asked by Eddie

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?
Answered by Henry

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 = ?


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