A 7.00-kg block of ice, released fromrest at the top of a 1.75-m-longfrictionles

Question

A 7.00-kg block of ice, released fromrest at the top of a 1.75-m-longfrictionless ramp, slides downhill, reaching a speed 2.25 m/s of at the bottom.
(a) What is the angle between the ramp and thehorizontal?


(b) What would be the speed of the ice at the bottom if the motionwere opposed by a constant friction force of 10.0 N parallel to thesurface of the ramp?
(a) What is the angle between the ramp and thehorizontal?


(b) What would be the speed of the ice at the bottom if the motionwere opposed by a constant friction force of 10.0 N parallel to thesurface of the ramp?

Explanation / Answer

m = 7.00-kg, s = 1.75-m, v = 2.25m/s a) find mgh = mv2/2 mg*(s*sin) = mv2/2 = arcsin(h/s) = arcsin[v2/(2gs)] =8.49o b) f = 10.0 N, find u mg*(s*sin) = mu2/2 + f*s u = [2s(gsin - f/m)] = 0.253 m/s

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