physics
posted by Eddie .
A 2.5kg 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?

physics 
Anonymous
22

physics 
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. FpFk = M*a.
7.1623.43u = 2.5*0.144.
u = ?
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