Posted by **Dee** on Wednesday, October 12, 2011 at 5:27pm.

Starting from rest, a 5.00 kg block slides 2.50 m down a rough 30.0 degree incline. The coefficient of kinetic friction between block and the incline is .436. Determine the work done by the friction force between block and incline and the work done on the normal force?

- Physics -
**Henry**, Friday, October 14, 2011 at 4:13pm
Wb = mg = 5kg * 9.8N/kg = 49N.

Fb = (49N,30deg.).

Fp = 49sin30 = 24.5N. = Force parallel to incline.

Fv = 49cos30 = 42.4N. = Force perpendicular to incline = The normal.

Ff = u*Fv = 0.436 * 42.4 = 18.50N. =

Force of friction.

1. Work=Ff * d = 18.50 * 2.5 = 46.3J.

2. d = h = 2.5 * sin30 = 1.25m.

Work = Fv * d = 42.4 * 1.25 = 53J.

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