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A 20-kg block slides down a frictionless incline from point A to point 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?

Respuesta :

The kinetic energy of the block at B is 24J

we can solve this by using law of conservation of energy,

Ki = 10J                     Uf = 0 ( placed at origin point B)

Kf=?

Ui =mgh

To find height, h = 2 sin 30 = 1m

from this Ui = mgh = 20x1 = 20 J

law of conservation of energy,

Ki+ Ui + W = Kf + Uf  - (1)

work = - F X D = 3X2 = - 6 N

substituting in equation 1

10+20- 6 = Kf + 0

30-6 = Kf

Kf = 24 J

hence, KE at B is 24J

To know more about friction,

https://brainly.com/question/13000653

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