An 8.0 kg block is released from rest, v1 = 0 m/s, on a rough incline, which has an angle of 40 from the horizontal. The block moves a distance of 1.6 m down the incline, in a time interval of 0.80 s, and acquires a velocity of v2 = 4.0 m/s.(adsbygoogle = window.adsbygoogle || []).push({});

The average rate at which friction force does work during the 0.80 s time interval is closest to:

a. +40 W-----------b. + 20 W ------------c. 0--------d. –40 W------------e. –20 W

From the force diagram that I sketched, I found the friction force to be F_fr = mu_k*mg*cos theta.

Then, a = g*sin(theta) – mu_k*g*cos(theta)

Mu_k = [a – g*sin(theta)]/[-g*cos(theta)] = (-1.2993/-7.50724) = 0.17307

Now find a.

v_f = v +a*t

a = (v_f-v)t = (4.0 m/s)/(0.80s) = 5.0 m/s^2

Now F_fr = mu_k*mg*cos theta = (0.17307)*(9.80 m/s^2)(8.0 kg)*cos 40

W of force = [mu_k*mg*cos theta]*d*cos(theta) = (0.17307)*(9.80 m/s^2)(8.0 kg)*cos 40*(1.6 m)*cos(180) = -16.631 J

P = -16.631 J/0.80 s = -20.788 W ??

Thanks.

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# Homework Help: Work on an Inclined Plane

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