Question

A 1.8 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 4.1 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 when the depth of the rock below the water's surface is 0 m.

Part B) Calculate the gravitational potential energy of the system, U, when the depth of the rock below the water's surface is 0 m.

Part C) Calculate the kinetic energy of the rock, K, when the depth of the rock below the water's surface is 0 m.

Part D) Calculate the total mechanical energy of the system, E, when the depth of the rock below the water's surface is 0 m.

Part E) Calculate the nonconservative work, Wnc, done by water resistance on the rock when the depth of the rock below the water's surface is 0.50 m.

Part F) Calculate the gravitational potential energy of the system, U, when the depth of the rock below the water's surface is 0.50 m.

Part G) Calculate the kinetic energy of the rock, K, when the depth of the rock below the water's surface is 0.50 m.

Part H) Calculate the total mechanical energy of the system, E, when the depth of the rock below the water's surface is 0.50 m.

Part I ) Calculate the nonconservative work, Wnc, done by water resistance on the rock when the depth of the rock below the water's surface is 1.0 m.

Part J)

Part A Part complete Calculate the nonconservative work, Wnc, done by water resistance on the rock when the depth of the rock below the water's surface is 0 m. Express your answers using two significant figures. Wnc = 0 J

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All attempts used; correct answer displayed

Part B

Part complete

Calculate the gravitational potential energy of the system, U, when the depth of the rock below the water's surface is 0 m.

Express your answers using two significant figures.

U =

32

J

Previous Answers

Correct

Part C

Calculate the kinetic energy of the rock, K, when the depth of the rock below the water's surface is 0 m.

Express your answers using two significant figures.

K K =

J

Previous AnswersRequest Answer

Incorrect; Try Again; 3 attempts remaining

Part D

Calculate 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.

E E =	nothing

J

Request Answer

Part E

Calculate the nonconservative work, Wnc, done by water resistance on the rock when the depth of the rock below the water's surface is 0.50 m.

Express your answers using two significant figures.

Wnc W n c =	nothing

J

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Part F

Calculate the gravitational potential energy of the system, U, when the depth of the rock below the water's surface is 0.50 m.

Express your answers using two significant figures.

U U =	nothing

J

Request Answer

Part G

Calculate the kinetic energy of the rock, K, when the depth of the rock below the water's surface is 0.50 m.

Express your answers using two significant figures.

K K =	nothing

J

Request Answer

Part H

Calculate 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.

E E =	nothing

J

Request Answer

Part I

Calculate the nonconservative work, Wnc, done by water resistance on the rock when the depth of the rock below the water's surface is 1.0 m.

Part J)Calculate the gravitational potential energy of the system, U, when the depth of the rock below the water's surface is 1.0 m.

Part K) Calculate the kinetic energy of the rock, K, when the depth of the rock below the water's surface is 1.0 m.

Part L) Calculate the total mechanical energy of the system, E, when the depth of the rock below the water's surface is 1.0 m.

Answer 1

Solution:

Given:

Mass (m) = 1.8 kg

Depth of pond (d) = 1.8 m

Resistive force (F) = 4.1 N (Upward)

Part (A) Solution: W_nc = F*x = (4.1 N) (0 m) = 0 J
Part (B) Solution: U = m g x = (1.8)(9.8)(1.8) = 31.752 J
Part (C) Solution: KE = 0 J (KE = U - W_nc )
Part (D) Solution: E = U + KE = 31.752 J
Part (E) Solution:  W_nc = F x = (4.1 N)( - 0.5 m) = - 2.05 J
Part (F) Solution: U = m g x = (1.8)(9.8)(1.3 m) = 22.932 J
Part (G) Solution: KE = 6.77 J  (KE = U - W_nc )
Part (H) Solution: E = U + KE = 29.702 J
Part (I) Solution:  W_nc = F*x = (4.1 N)(-1m) = - 4.1 J
Part (A) Solution: U = m g x = (1.8)(9.8)(0.8m) = 14.122 J
Part (J) Solution: KE = 4.71 J
Part (K) Solution: E = U + KE = 18.832 J