In a truck-loading station at a post office, a small 0.200-kg package is release

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 truck that is one-quarter of a circle with radius 1.60m (the figure (Figure 1)). The size of the package is much less than 1.60 mm, so the package can be treated as a particle. It slides down the track and reaches point B with a speed of 4.50 m/s. From point B, it slides on a lower surface a distance of 3.00 m to point C, where it comes to rest. What is the coefficient of kinetic friction on the horizontal surface? mu = How much work is done on the package by friction as it slides down the circular are from A to B? W =

Explanation / Answer

Part A

Let deceleration of block be a

horizontal distance travelled by block before stopping=s=3m

Velocity of block at B=u=4.5m/s

Velocity of block at C=v=0

v²=u²+2as

0=4.5²+2*a*3

a=-3.375m/s²

Let coefficient of kinetic friction be f

f=3.375/9.8 =0.34

Part B

Inital potential energy of block at A= mgR=0.2*9.8*1.6=3.136 Joules

Kinetic energy of block at B=(1/2)mv²=(1/2)*0.2*4.5²=2.025Joules

Work done by frictional force on block in moving from A to B=loss in energy from A to B=3.136-2.025=1.111 Joules

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