A horizontal force of 80 N acts on a mass of 6 kg resting on a horizontal surfac

Question

A horizontal force of 80 N acts on a mass of 6 kg resting on a horizontal

surface.

The mass is initially at rest and covers a distance of 5 m in 0.92 s

under the action of the force.

Assuming there are no energy losses due to

air resistance and therefore that the acceleration is constant:

Calculate the total energy expended in the acceleration.

Plot a graph of the Kinetic energy of the mass against time

Plot a graph of the kinetic energy of the mass against distance

Calculate the coefficient of friction between the mass and the surface

Explanation / Answer

If there were no friction, teh force F would do work W given by:

W = Fd where d = distance the force acts over so using your numbers

W = 80N*5m = 400 J (a)

Now it takes the box 0.92s to move the 5m and you are told acceleration is constant so you can use

d = 1/2 at^2 to find a ---> a = 2d/t^2 = 2*5 m/(0.92s)^2 = 11.82 m/s^2

The net force, the difference between the applied force F and friction, is then:

Fnet = ma = 70.92 =N

Then,

Fnet = F - umg where u = coefficient of friction and g = 9.8 m/s^2

SOlving for u ---> u = (F - Fnet)/mg = 0.154

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