In a truck-loading station at a post office, a small 0.200- k g package is relea

Question

In a truck-loading station at a post office, a small 0.200-kg package is released from rest at point A on a track that is one-quarter of a circle with radius 1.60 m. The size of the package is much less than 1.60 m, so the package can be treated as a particle. It slides down the track and reaches point Bwith a speed of 4.20m/s . From point B, it slides on a level surface a distance of 3.00 m to point C, where it comes to rest.

A.What is the coefficient of kinetic friction on the horizontal surface?

B.How much work is done on the package by friction as it slides down the circular arc from

A to B?

Explanation / Answer

Calc the KE at the bottom of the slope KE = 1/2 mv^2 = 1/2 .2*(4.4)^2 = 1.94 J

Now work off that KE with friction work KE = 1.94 = k N d = k mg d = k .2*9.81*3; so that k = 1.94/5.886 = .33. N is the normal weight = mg of the package and d is the stopping distance once it reaches the horizontal. WE = kNd is the work function and kN is the friction force that's braking the package and bringing it to a halt.

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