Question

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, *W*nc , 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.

*W*nc*a* , *U**a* ,
*K**a* , *E**a=________J*

Part B

Calculate the nonconservative work, *W*nc , 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

*W*nc*b* , *U**b* ,
*K**b* , *E**b* =________J

Part C

Calculate the nonconservative work, *W*nc , 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.

*W*nc*c* , *U**c* ,
*K**c* , *E**c* =_______J

Answer #1

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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