How Does Kinetic Energy Change on an Inclined Plane with Friction?

AI Thread Summary
The discussion focuses on a physics problem involving a crate on an inclined plane with friction. Key calculations include determining the increase in internal energy due to friction, the change in kinetic energy, and the final speed of the crate after being pulled 4.92 meters. The relevant equations include kinetic energy (KE = 0.5mv^2), the frictional force, and work done. The total energy of the system remains constant, with energy being redistributed among kinetic energy, potential energy, and internal energy lost to friction. Understanding the relationship between these energy forms is crucial for solving the problem effectively.
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Homework Statement



A crate of mass 10.2 kg is pulled up a rough incline with an initial speed of 1.42 m/s. The pulling force is 94.0 N parallel to the incline, which makes an angle of 20.7° with the horizontal. The coefficient of kinetic friction is 0.400, and the crate is pulled 4.92 m.

(b) Determine the increase in internal energy of the crate-incline system due friction.
(d) What is the change in kinetic energy of the crate?
(e) What is the speed of the crate after being pulled 4.92 m?

Homework Equations



KE = .5mv^2
frictional force = u(Fn)
work = F x distance

The Attempt at a Solution



for b, i thought it should just be the force of friction x distance so i did uFn(4.92)
for the others i don't really know
 
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The total energy of the system is conserved, that is, E=K+U+E_{int}=constant.
In general, K=1/2mv^2, U=mgy and E_{int} is the energy lost to friction.

Draw a picture of the initial and final states of the system, and keep in mind that the energy is the same in both states, though K, U, and E_{int} may have different values.
 
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