A 1.6kg rock is released from rest at the surface of a pond 1.8 m deep. As the r

Question

A 1.6kg rock is released from rest at the surface of a pond 1.8 m deep. As the rock falls, a constant upward force of 4.8N is exerted on it by water resistance. Let y=0 be at the bottom of the pond.

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 .

Express your answers using two significant figures separated by commas.

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 .

Express your answers using two significant figures separated by commas.

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 .

Express your answers using two significant figures separated by commas

Explanation / Answer

nonconservative work, Wnc = -4.8 N x d    ( d = displacement of rock below water)

gravitational potential energy of the system, U = mgh (height ''h''from from bottom y=0)

the kinetic energy of the rock, K = (0.5) m V^2

mechanical energy of the system, E = U +K +Wnc

a) d =0   , h = 1.8 m , m = 1.6 kg

nonconservative work, Wnc = -4.8 N x d = -4.8 N x 0 =0

gravitational potential energy of the system, U = mgh = 1.6*9.8*1.8 = 28.224 J

the kinetic energy of the rock, K = (0.5) m V^2 = (0.5) m 0^2 =0 J

mechanical energy of the system, E = U +K +Wnc = 28.224 J +0 +0 = 28.224 J

b) d = 0.50 m , h = 1.8 - 0.5 = 1.3 m

nonconservative work, Wnc = -4.8 N x 0.5 = -2.4 N

gravitational potential energy of the system, U = mgh = 1.6*9.8*1.3 = 20.38 J

mechanical energy of the system, E = 28.224 J           from conservation of energy

the kinetic energy of the rock, K = E- U - Wnc = 28.224 - 20.28 - (-2.4) = 10.34 J

c) d = 1 m , h = 1.8 - 1 = 0.8 m

nonconservative work, Wnc = -4.8 N x 1 = -4.8 N

gravitational potential energy of the system, U = mgh = 1.6*9.8*0.8 = 12.54 J

mechanical energy of the system, E = 28.224 J           from conservation of energy

the kinetic energy of the rock, K = E- U - Wnc = 28.224 - 12.54 - (-4.8) = 20.48 J

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