A 17-kg sled is being pulled along the horizontal snow-covered ground by a horizontal force of 27 N. Starting from rest, the sled attains a speed of 2.4 m/s in 8.3 m. Find the coefficient of kinetic friction between the runners of the sled and the snow.

The coefficient of kinetic friction between the runners of the sled and the snow is μ = 0.12

Explanation:

The work energy theorem, change in kinetic energy of the object  from initial position to the final position is equal to the work done on the object ie when the force is applied on  the object the object changes its position and work is done on the object.

The work done is given by W =Fdcosθ,

     where F is the force,

                d is the displacement,

                θ is angle between force and displacement.

The work done on the object is equal to difference between the final and initial kinetic energy.

W = K.E₂  -  K.E ₁

where, K.E₂ is final kinetic energy,

             K.E ₁ is initial kinetic energy.

Given are mass of sled m = 17 kg,

                 Pulling force F = 27N,

                 final velocity v₂ = 2.4m/s,

                 displacement d = 8.3m.

The work done by sled to attain to attain final velocity W = K.E₂ = 0.5mv²₂

     W =0.5*17*(2.4)²

       W = 48.96J.

So, the work done by the sled to final velocity is 48.96J.

The net work by the sled is some of the work done by pulling force and work done by the friction due to snow covered ground.

W = W₁ +W₂

As the pulling force acts along the direction of motion the angle between the force and displacement is zero.

So the work done by the pulling force is W₁ = Fd cos0 = 27*8.3 = 224.1 J.

The work done by the friction force is W₂ =W-W₁ = 48.96 - 224.1 = - 175.14 J.

The frictional force acts in the direction opposite to that of motion hence the angle between force and displacement is θ = 180°.

     W₂ =F d cos180°

    - 175.14J. =f *8.3*9*(-1)

f = =21.1 N

The frictional force is f =21.1N

The frictional force is give by f = μN

Where N is the normal force on the sled, μ is coefficient of kinetic friction.

There is no motion along vertical direction and hence acceleration along this direction is zero.

Applying Newton second law of motion along the vertical direction,

     F = N - mg = 0

N = mg = 17*9.8 = 166.6 N

So the coefficient of kinetic friction between the runners of the sled and the snow is

μ = = =  0.12

The coefficient of kinetic friction between the runners of the sled and the snow is μ = 0.12

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another example of kinetic energy would be,

kinetic energy   >     moving energy

answer   >   letter choice (c),   a hammer falling off a table.

hope that !                                         : )