A 1.65-kg rock is released from rest at the surface of a pond 1.00 m deep. As th
ID: 2056544 • Letter: A
Question
A 1.65-kg rock is released from rest at the surface of a pond 1.00 m deep. As the rock falls, a constant upward force of 4.40 N is exerted on it by water resistance. Calculate the nonconservative work, Wnc, done by water resistance on the rock, the gravitational potential energy of the system, U, the kinetic energy of the rock, K, and the total mechanical energy of the system, E, for the following depths below the water's surface. Let y = 0 be at the bottom of the pond.(a) d = 0.00 m
(b) d = 0.50 m
(c) d = 1.00 m
For each distance please find the:
Wnc done by water resistance on the rock
gravitational potential energy of the system
kinetic energy of the rock
total mechanical energy of the system
Explanation / Answer
Non conservative Work will be the nonconservative force (4.4) times the distance it moved.
The gravitational PE will be found by mgh above the floor of the pond
KE can be found the conservation of energy. KE will be found by the total energy possible (Which is the total PE at the top of the surface) minus how much nonconservative work + PE it has at the individual points.
Total mechanical energy is the sume of PE + KE
At point A)
W = Fd, since d is zero, W = 0J
PE = mgh = (1.65)(9.8)(1) = 16.17 J
KE will be zero at this point since it has not started falling yet. It is at the surface
Total ME = PE + KE = 16.17 J
At point B)
W = Fd = (4.4)(.5) = 2.2 J
PE = mgh = (1.65)(9.8)(.5) = 8.085 J
KE = Total E - PE - W = 16.17 - 2.2 - 8.085 = 5.885 J
Total ME = PE + KE = (8.085) + (5.885) = 13.97 J
At point C)
W = Fd = (4.4)(1) = 4.4 J
PE = zero since it is now at the bottom
KE = 16.17 - 4.4 = 11.77 J
ME = PE + KE = 0 + 11.77 = 11.77 J
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