A 2.5 kg rock is released from rest at the surface of a pond 1.8 m deep. As the

Question

A 2.5 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 4.2 N is exerted on it by water resistance. Let y=0 be at the bottom of the pond. Calculate the nonconservative work, Wnc, done by water resistance on the rock, the gravitational potential energy of the system, PE, the kinetic energy of the rock, KE, and the total mechanical energy of the system, E, when the depth of the rock below the water's surface is 0.50m .

How do I get the KE? I'm stuck on figuring out the velocity to plug in to PE=.5mv^2. And would I need to subtract the Wnc?

So far I have:

Wnc= fd= (4.2N)(1.3m)= 5.46J

PE= mgh= (2.5kg)(9.81m/s^2)(1.3m)= 31.85J

Explanation / Answer

Wnc= fd= (4.2N)(-0.5)=- 2.1J PE= mgh= (2.5kg)(9.8m/s^2)(1.3m)= 31.85J KE=change in potential energy-loss of energy=[2.5*9.8*(1.8-1.3)-2.1]=10.15 mechanical energy of the system, E=KE+PE=42 J

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