How Does Friction Impact Mechanical Energy on an Incline?

[Haroon Pasha]

A crate of mass 10.0 kg is pulled up a rough incline with an initial speed of 1.50 m/s. The pulling force is 100 N parallel to the incline, which makes an angle of 20 degrees with the horizontal. The coefficient of kinetic friction is 0.400, and the crate is pulled 5.00 m. How much mechanical energy is lost due to friction and what is the change in kinetic energy of the crate?

I figured out the kinetic energy initially, and i got 11.25 J. For the final kinetic energy i got 501.00 J(using the equation KE=1/2*m*v2(squared).

How do i find the change in kinetic energy? do I subtract the final from initial ? and for the mechanical energy lost doing friction, how does the friction fit into the equation?

 

[dextercioby]

How did u get that 501J...?HINT:use the theorem of variation of KE.How many forces act on the crate and how many of them do work on the crate...?

Daniel.

 

[thepatientmental]

To find the change in kinetic energy, you would subtract the final kinetic energy from the initial kinetic energy. So in this case, the change in kinetic energy would be 501.00 J - 11.25 J = 489.75 J. This means that the crate's kinetic energy increased by 489.75 J as it was pulled up the incline.

To calculate the mechanical energy lost due to friction, you would need to use the formula for work: W = Fdcosθ. In this case, the force of friction is equal to the coefficient of kinetic friction (0.400) multiplied by the normal force (mgcosθ). The normal force is equal to the weight of the crate, which is mg, and cosθ is equal to the cosine of the angle between the incline and the horizontal (in this case, 20 degrees). So, the formula for work would be W = (0.400)(mgcosθ)(d).

To find the distance (d), we can use the Pythagorean theorem to calculate the hypotenuse of the triangle formed by the incline and the horizontal surface. So, d = √(5.00^2 + 1.50^2) = √26.25 = 5.123 m.

Plugging in all the values, we get W = (0.400)(10.0 kg)(9.8 m/s^2)(cos20°)(5.123 m) = 20.04 J. This means that the mechanical energy lost due to friction is 20.04 J.

To summarize, the change in kinetic energy of the crate is 489.75 J and the mechanical energy lost due to friction is 20.04 J. Both of these values are positive, indicating that the crate gained kinetic energy and lost mechanical energy as it was pulled up the incline.