The figure shows a 0.426-kg block sliding from A to B along a frictionless surfa

Question

The figure shows a 0.426-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 11.4 and 7.30 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

Here ,

mass , m = 0.426 Kg

hA = 11.4 m

hB = 7.30 m

hC = 7.30 m

a) at point C

Using conservation of energy , Kinetic energy at B

Kinetic energy at B = m *g * (hA - hB)

Kinetic energy at B = 9.8 * 0.426 * (11.4 - 7.30)

Kinetic energy at B = 17.12 J

b)

as the kinetic energy lost due to frictional force

kinetic energy lost due to frictional force = kinetic energy at B

kinetic energy lost due to frictional force = 17.12 J

the kinetic energy lost due to frictional force is 17.12 J

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