Calculating Work and Friction on a Crate on an Incline

AI Thread Summary
The discussion revolves around calculating the work done by gravitational force on a crate being pulled up an incline and the increase in internal energy due to friction. The crate has a mass of 10.6 kg, is pulled with a force of 104 N at an angle of 20.1°, and experiences a coefficient of kinetic friction of 0.400. The user has successfully calculated the gravitational work as -180 J but seeks clarification on how to determine the increase in internal energy related to friction. It is suggested that understanding the work done by friction, which contributes to heating the crate, is essential for solving part B of the problem. The conversation emphasizes the relationship between work, friction, and internal energy in this context.
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Homework Statement


A crate of mass 10.6 kg is pulled up a rough incline with an initial speed of 1.48 m/s. The pulling force is 104 N parallel to the incline, which makes an angle of 20.1° with the horizontal. The coefficient of kinetic friction is 0.400, and the crate is pulled 5.05 m.

(a) How much work is done by the gravitational force on the crate?
(b) Determine the increase in internal energy of the crate-incline system due friction.

The Attempt at a Solution



I have solved part A, and I got -180 J. I need help on B. I really don't know what the question is looking for. Thanks
 
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Increase in internal energy is decrease in kinetic energy.
 
It may be easier to think about it in terms of the definition of work done by a force. That is,

W = \int Fds

where the work done by friction rubs the molecules beneath the crate and heats it up, hence an increase in internal energy.
 
So do I need to find the Work that is done by friction?
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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