Two questions, work-kinetic energy related

[BifSlamkovich]

Homework Statement

1)A 5.0-kg block slides down a surface inclined 30 degrees with the horizontal. If the coefficient of kinetic friction for the block and the surface is equal to 0.25, at what rate is the friction force on the block doing work at an instant when the speed of the block is 3.0 m/s?


2) Consideran incline with two locations on the slope, A and B. point A on the incline does not necessarily occur at the top of the incline but does occur before B.
A 2.0-kg block slides down a frictionless incline from p.A to p.B. A force (magnitude P = 3.0 N) acts on the block between A and B, as shown. Points A and B are 2.0 m apart. If the kinetic energy of the block at A is 10 J, what is the kinetic energy of the block at B?

Homework Equations

The Attempt at a Solution

i keep getting incorrect numbers (the questions are multiple choice).

 

[Nate810]

For the first one I used P=Fk and W=F*Δx but i dont know what i'm doing wrong. For the second one, i used KEf=KEi+WAB but again, wrong answer. Please help!

 

[Moetasim]

1) The rate of work done by friction can be calculated using the formula W = Fd, where W is work, F is force, and d is distance. In this case, the force of friction is equal to the coefficient of kinetic friction (µ) multiplied by the normal force (N) exerted on the block. The normal force can be found using the formula N = mgcosθ, where m is the mass of the block, g is the acceleration due to gravity, and θ is the angle of inclination.

Therefore, the rate of work done by friction can be calculated as follows:

W = µNcosθ * v

Where v is the velocity of the block. Substituting the given values, we get:

W = (0.25)(5.0 kg)(9.8 m/s^2)cos30 * 3.0 m/s

W = 10.3 J/s

Therefore, the rate of work done by friction on the block at that instant is 10.3 J/s.

2) The kinetic energy of the block at point A is given as 10 J. At point B, the block is still moving and has kinetic energy. This energy is equal to the work done on the block by the force P between points A and B.

Using the work-energy theorem, we can calculate the work done by the force P as:

W = Fd = Pd

Where d is the distance between points A and B, which is given as 2.0 m.

Therefore, the work done by the force P is 3.0 N * 2.0 m = 6.0 J.

Since the kinetic energy at B is equal to the work done on the block, the kinetic energy at B is also 6.0 J.