Suggested E.1 In the figure, a 3-kg block is sent through point A with a kinetic

Question

Suggested E.1 In the figure, a 3-kg block is sent through point A with a kinetic energy of 74 J. Its path is without friction until it reaches the section of length L = 16 m. When the particle reaches point D, its kinetic energy is 8 J. The indicated heights are h_1 = 7 m and h_2 = 2 m. m. a. What is the kinetic energy of the block at point B? 284 J b. What is the kinetic energy at point C? 224 J c. What is the work done by friction when the block slides from point A to D? -216 J d. What is the frictional force on the block when it slides through the section of length L? 13.5 N

Explanation / Answer

(a) Initial energy of the block at point A = kinetic energy + potential energy
= 74 + mgh1 = 74 + 3*10*7 = 284 J
Energy at point B = potential energy + kinetic enery = 0 + Kinetic energy
Since energy always remain constant therefore
Energy at point A = energy at point B
Kinetc energy at point B = 284 J
(b) Energy at point C = kinetic energy + potential energy
Kinetic energy + (mgh2) = kinetic energy + (3*10*2)
Energy at point C must be equal to the energy at point B
kinetic energy + (3*10*2) = 284
Kinetic energy = 284 - 60 = 224 J
(c) At point D kinetic energy is 8 J
So if we take energy conservation between point C and D
since they are at same height thereofre potential energy can be neglected.
Energy At C = energy at D
224 = kinetic energy + frictional energy
224 = 8 + frictional energy
frictional energy = 8 -224 = -216 J
that means energy is waste or spend in overcoming friction .
(d) frictional energy = friction force*distance
216 = friction force*16
friction force = 13.5 N

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