# Physics

A 3.0 kg block is moved up a 37 degree incline under a constant horizontal force of 40.0 N. The coefficient of friction is 0.10 and the block is displaced 2.0 m up the incline. Calculate the change in the kinetic energy of the block.

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asked by Andrew
1. Wb = m*g = 3kg * 9.8N/kg = 29.4 N. =
Weight of block.

Fp = 29.4*sin37 = 17.69 N. = Force
parallel to the incline.

Fn = 29.4*cos37 = 25.61 N. = Normal force = Force perpendicular to the
incline.

Fk = u*Fn = 0.1 * 25.61 = 2.56 N =

!@#\$%^& = 40/cos37 = 50.09 N. Parallel to
incline.

a = (!@#\$%^&-Fk)/m = (50.09-2.56)/3 = 15.84
m/s^2.

V^2 = Vo^2 + 2a*d = 0 + 2*15.84*2=63.4
V = 7.96 m/s.

KE=0.5m*V^2 = 0.5*3*63.4 = 95.1 Joules.

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posted by Henry
2. Correction:

a = (!@#\$%^&-Fp-Fk)/m = (50.09-17.69-2.56)/3
= 9.95 m/s^2.

V^2 = Vo^2 + 2a*d = 0 + 2*9.95*2 = 39.8
V = 6.30 m/s.

KE = 0.5m*V^2 = 0.5*3*39.8 = 59.7 J.

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posted by Henry

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