Question

# A 2.1 kg rock is released from rest at the surface of a pond 1.8 m...

A 2.1 kg rock is released from rest at the surface of a pond 1.8 m deep. As the rock falls, a constant upward force of 4.9 N is exerted on it by water resistance. Let y=0 be at the bottom of the pond.

Part A

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 , when the depth of the rock below the water's surface is 0 m .

Wnca , Ua , Ka , Ea=________J

Part B

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 , when the depth of the rock below the water's surface is 0.50 m .

Wncb , Ub , Kb , Eb =________J

Part C

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 , when the depth of the rock below the water's surface is 1.0 m .

Wncc , Uc , Kc , Ec =_______J

a) rock at 0 depth
Work by water resistance = W(1.8) = 0
P.E(1.8). = mgh = (2.1)(9.8)(1.8) = 37.044 J
K.E(1.8). = 1/2mv² = 0
Total M.E. of system = P.E. = 37.044 J

b) rock at 0.50 m depth
Work by water resistance W(1.3) = F x d = (4.9)(0.50) = 2.45 J
P.E(1.3). = mgh = (2.1)(9.8)(1.8-0.5) = (20.58)(1.3) = 26.754 J
K.E(1.3). = Total M.E. - P.E.(1.3) - W(1.3)
K.E.(1.3) = 37.044 - 26.754 - 2.45 = 7.84 J
Total M.E. of system = 37.044 J (constant)

c) rock at 1.0 m depth
Work by water resistance W(0.8) = F x d = (4.9)(1.0) = 4.9 J
P.E.(0.8) = mgh = (2.1)(9.8)(1.8 - 1.0) = (20.58)(0.8) = 16.464 J
K.E.(0.8) = Total M.E. - P.E.(0.8) - W(0.8)
K.E.(0.8) = 37.044 - 16.464 - 4.9 = 15.68 J
Total M.E. of system = 37.044 J (constant)

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