How do you do this work and energy problem? physics urgent help!?
A 4.5 kg box slides down a 4.4 m high frictionless hill, starting from rest. It then travels across a 2.4 m wide horizontal surface before hitting a horizontal spring with a spring constant of 500 N/m. The other end of the spring is anchored against a wall. The ground beneath the spring is frictionless, but the 2.4 m long horizontal surface is rough, with a coefficient of kinetic friction of 0.27.
What is the speed of the box just before it reaches the rough surface? What is the speed of the box just before it hits the spring? How far is the spring compressed? Including the first crossing, how many complete trips will the box make across the rough surface before coming to rest?
1 Answers
Energy balance with loss due to friction
at top of frictionless hill the box has its TOTAL starting energy as P.E.
P.E. top = mgh {where m = 4.5, g = 9.8, h = 4.4)
P.E. top = (4.5)(9.8)(4.4) = 194 J {total energy at start}
at bottom of frictionless hill the box has its TOTAL energy as K.E.
P.E. top = K.E. bottom = 1/2 mv²
194 = 0.5(4.5)v²
v² = 194/2.25 = 86.22
v = 9.29 m/s ANS-1
over rough surface friction force extracts work from box energy total
Wf = FF x 2.4 {where Wf = work done by friction force of FF)
FF = (0.27) x normal force
normal force = box weight = mg = (4.5)(9.8) = 44.1 N
FF = (0.27)(44.1) = 11.9 N
Wf = (11.9)(2.4) = 28.6 J
box speed just before hitting spring is given by K.E. of box
K.E. before spring = K.E.bottom - Wf = 194 - 28.6 = 165.4 J
K.E. before spring = 1/2 mv²
165.4 = (0.5)(4.5)v²
v² = 165.4/2.25 = 73.51
v = 8.57 m/s ANS-2
K.E. box = 165.4 J is stored at max spring compression as P.E. spring
165.4 = P.E. spring = 1/2 kx²
165.4 = (0.5)(500)x²
x² = 165.4/250 = 0.662
x = 0.81 m ANS-3
Each time box crosses rough surface it loses Wf = 28.6 J
Box starts with a total energy = 194 J
Box can cross rough surface = 194/28.6 = 6.8 times
since question asks for "complete trips" answer is 6 times ANS-4
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