A 2.4 kg rock is released from rest at the surface of a pond 1.8 m deep. As the
ID: 1544214 • Letter: A
Question
A 2.4 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 5.0 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. 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. 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.
Explanation / Answer
a)
rock at 0 depth
Work by water resistance = W(1.8) = 0
P.E(1.8). = mgh = (2.4)(9.81)(1.8) = 42.38 J
K.E(1.8). = 1/2mv^2 = 0
Total M.E. of system = P.E. = 42.38 J
b)
rock at 0.50 m depth
Work by water resistance W(1.3) = F x d = (5)(0.50) = 2.5 J
P.E(1.3). = mgh = (2.4)(9.81)(1.8-0.5) = 30.6 J
K.E(1.3). = Total M.E. - P.E.(1.3) - W(1.3)
K.E.(1.3) = 42.38 - 30.6 - 2.5 = 9.28 J
Total M.E. of system = 39.88 J
c)
rock at 1.0 m depth
Work by water resistance W(0.8) = F x d = (5)(1.0) = 5 J
P.E.(0.8) = mgh = (2.4)(9.81)(1.8 - 1.0) = 18.84 J
K.E.(0.8) = Total M.E. - P.E.(0.8) - W(0.8)
K.E.(0.8) = 42.38 - 18.84 - 5 = 18.54 J
Total M.E. of system = 37.38 J
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.