1) A boy releases his 2.0 kg toy, from rest, at the top of a sliding-pond inclin

Question

1) A boy releases his 2.0 kg toy, from rest, at the top of a sliding-pond inclined at 20o above the horizontal. What will the toy's speed be after sliding 4.0 m along the sliding-pond? The coefficient of kinetic friction is 0.20. 2) A race car, traveling at a constant speed of 50 m/s, drives around a circular track of radius 250 m. What angular acceleration does it experience? 1) A boy releases his 2.0 kg toy, from rest, at the top of a sliding-pond inclined at 20o above the horizontal. What will the toy's speed be after sliding 4.0 m along the sliding-pond? The coefficient of kinetic friction is 0.20. 2) A race car, traveling at a constant speed of 50 m/s, drives around a circular track of radius 250 m. What angular acceleration does it experience?

Explanation / Answer

PE1 + KE1 = PE2 + KE2 + Elost, where PE = mgh, and KE= .5mv2 if we place our datum (origin at the 4m mark) PE2 = 0 and since it starts from rest KE1 = 0. Elost is the energy dissipated, due to friction ( Elost = fd, friction force times the horizontal distance). From the FRB you can see that N=mgcos() and f=N, where N is the normal force. Putting this together yields, gh1 = 0.5v22 + dgcos, where = fric. coeff. d=horizontal distance traveled and is the angle made with the horizon. solve for v2 for # 2 ang. acc = r2 , v=r, ang. acc.=v2/r where is the angular velocity, r= radius and v= velocity. Hope this helps.

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