Physics help? Topic: kinetic energy?

[oligosaccharide]

A block of mass 10 kg is pulled 5 m upwards along a rough surface at an angle of 20 deg, an external force of 100 N and vi = 1.5 m/s. The coefficient of friction is 0.4. Calculate:

1) The work done by gravity
2) The increase in internal energy between the block and the surface as a result of friction
3) Work done by external force
4) The change in kinetic energy of the block
5) The speed of the block after moving 5 m.

Thanks a lot!

 

[gezibash]

Please show your work first, and then we will be able to guide you.

 

[oligosaccharide]

For 1), I calculated the work done by gravity by musing the formula Fd, so Wg = mgsintheta x d, so Wg = 167.6 J.

I don't know how to do 2)

For 3), I multiplied the external force by cos 20 and then by 5 m = 469.8 J

For 4), the change in kinetic energy equals the work done by external force minus the work done by friction.
The frictional force is the coefficient of friction multiplied by the normal force (N = mg - F sin theta).
So the frictional force is 25.5 N
Change in kinetic energy = 469.8 J - (25.5 x 5) J = 342.4 J

For 5), change in kinetic energy = 1/2 x m(vf^2 - vi^2)
And I got vf = 8.36 m/s

 

[BvU]

Hello sugar and welcome to PF. Good thing you re-posted with at least something. Next time use the template. Read the guidelines to find out why that's reall really useful.

2) might have something to do with the work the friction force 'does'.
3) Does the external force pull along the slope or does it pull horizontally ? Or vertically ?
4) Not correct. you forget another energy change. And, again, which way is F pulling in your N calculation ?

And: make a drawing !

 

[HalfLostAndSearching]

1) The work done by gravity is given by the equation W = mgh, where m is the mass of the block, g is the acceleration due to gravity, and h is the height the block is pulled. In this case, the block is pulled upwards by 5 m, so the work done by gravity is (10 kg)(9.8 m/s^2)(5 m) = 490 J.

2) The increase in internal energy between the block and the surface as a result of friction can be calculated using the equation W = µNΔx, where µ is the coefficient of friction, N is the normal force, and Δx is the distance the block moves. The normal force is equal to the weight of the block, which is (10 kg)(9.8 m/s^2) = 98 N. So, the increase in internal energy is (0.4)(98 N)(5 m) = 196 J.

3) The work done by the external force is given by the equation W = FΔx, where F is the external force and Δx is the distance the block moves. In this case, the external force is 100 N and the block moves 5 m, so the work done by the external force is (100 N)(5 m) = 500 J.

4) The change in kinetic energy of the block can be calculated using the equation ΔKE = ½mvf^2 - ½mvi^2, where m is the mass of the block, vf is the final velocity, and vi is the initial velocity. In this case, the mass is 10 kg, the initial velocity is 1.5 m/s, and the final velocity can be found using the equation vf^2 = vi^2 + 2aΔx, where a is the acceleration of the block. The acceleration can be found using Newton's second law, Fnet = ma, where Fnet is the net force acting on the block. In this case, the net force is equal to the external force minus the force of friction, so Fnet = (100 N - 0.4(98 N)) = 60.8 N. Therefore, the acceleration is a = (60.8 N)/(10 kg) = 6.08 m/s^2. Plugging in these values, we get ΔKE = ½(10 kg)(vf^