Preston pushes a wheelbarrow weighing 500 N to the top of a 50.0-m ramp,inclined at 20.0 $\degree$ with the horizontal,and leaves it.Tamara accidentally bumps the wheelbarrow.It slides back down the ramp,during which an 80.0-N frictional force acts on it over the 50.0 m.What is the wheelbarrow's kinetic energy at the bottom at of the ramp? (g = 9.8 m/s²) A)4 550 J B)6 550 J C)8 150 J D)13 100 J

Question

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Explanation:The work done by Preston in pushing the wheelbarrow up the ramp is:$$W = Fd\cos	heta = (500	ext{ N})(50.0	ext{ m})\cos20.0^\circ = 45.5	ext{ kJ}$$The work done by friction in sliding back down the ramp is:$$W = Fd = (80.0	ext{ N})(50.0	ext{ m}) = 4.00	ext{ kJ}$$The net work done on the wheelbarrow is:$$W = 45.5	ext{ kJ} - 4.00	ext{ kJ} = 41.5	ext{ kJ}$$The kinetic energy of the wheelbarrow at the bottom of the ramp is:$$K = W = 41.5	ext{ kJ} = \boxed{4150	ext{ J}}$$

Preston Pushes a Wheelbarrow Weighing 500 N to the Top °\degree°

Preston Pushes a Wheelbarrow Weighing 500 N to the Top $\degree$