The figure shows a 0.388-kg block sliding from A to B along a frictionless surfa

Question

The figure shows a 0.388-kg block sliding from A to B along a frictionless surface. When the block reaches B, it continues to slide along the horizontal surface BC where the kinetic frictional force acts. As a result, the block slows down, coming to rest at C. The kinetic energy of the block at A is 43.0 J, and the heights of A and B are 10.2 and 6.30 m above the ground, respectively. (a) What is the value of the kinetic energy of the block when it reaches B? (b) How much work does the kinetic frictional force do during the BC segment of the trip?

Explanation / Answer

a)

here from energy conservation

KEb + m * g * hb = KEa + m * g * ha

KEb = KEa + m * g * ( ha - hb)

KEb = 43 + 0.388 * 9.8 * ( 10.2 - 6.3)

KEb = 57.83 J

b)

there is non conservative force

Wnc = KEc + m * g * hc - ( KEb + m * g * hb)

Wnc = KEc - KEb + m * g * (hc - hb)

Wnc = -KEb = -57.83 J

then

for BC trip there is no total energy conserved

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