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

[ramenmeal]

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

 

[BAnders1]

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.

 

[nlightner]

where to start.



I would approach this problem by first defining the terms and variables involved. The kinetic energy of a crate refers to the energy it possesses due to its motion. It is calculated using the equation KE = 0.5mv^2, where m is the mass of the crate and v is its velocity. In this case, the crate has a mass of 10.2 kg and an initial speed of 1.42 m/s.

The problem also mentions a pulling force of 94.0 N, which is the force applied to the crate parallel to the incline. This force is making an angle of 20.7° with the horizontal. The coefficient of kinetic friction is given as 0.400, which is a measure of the roughness between the crate and the incline.

For part (b), we need to determine the increase in internal energy of the crate-incline system due to friction. This can be calculated using the formula Work = Force x Distance, where the work done by friction will be equal to the frictional force (uFn) multiplied by the distance the crate is pulled (4.92 m).

For part (d), we need to find the change in kinetic energy of the crate. This can be calculated by subtracting the initial kinetic energy from the final kinetic energy. The initial kinetic energy can be calculated using the given mass and initial velocity, while the final kinetic energy can be calculated using the final velocity, which we will determine in part (e).

For part (e), we need to find the speed of the crate after being pulled 4.92 m. This can be calculated using the equation Work = Change in Kinetic Energy, where the work done by the pulling force will be equal to the change in kinetic energy of the crate. We can solve for the final velocity using this equation, and then use this value to calculate the final kinetic energy in part (d).

In summary, to solve this problem, we need to use the equations for kinetic energy, frictional force, work, and change in kinetic energy. We also need to carefully consider the given variables and their relationships to each other. I hope this helps.