A sled plus passenger with total mass m = 53.1 kg is pulled a distance d = 25.3 m across a horizontal, snow-packed surface for which the coefficient of kinetic friction with the sled is μk = 0.155. The pulling force is constant and makes an angle of φ = 28.3 degrees above horizontal. The sled moves at constant velocity.

Required:
a. Write an expression for the work done by the pulling force in terms of m, g (acceleration due to gravity), φ, μk, and d.
b. What is the work done by the pulling force, in joules?
c. Write an expression for the work done on the sled by friction in terms of m, g (acceleration due to gravity), φ, μk, and d.
d. What is the work done on the sled by friction, in joules?

Explanation:

Force of friction

F = μ mg

μ is coefficient of friction , m is mass and g is acceleration due to gravity .

If f be the force applied to pull the sled , the horizontal component of force should be equal to frictional  force

The vertical component of applied force will reduce the normal force or reaction force from the ground

Reaction force R = mg - f sin28.3

frictional force = μ R where μ is coefficient of friction

frictional force = μ x (mg - f sin28.3 )

This force should be equal to horizontal component of f

μ x (mg - f sin28.3 ) = f cos 28.3

μ x mg = f μsin28.3 + f cos 28.3

f = μ x mg / (μsin28.3 + cos 28.3 )

a )

work done by pulling force  = force x displacement

f cos28.3 x d

μ x mg d cos28.3  / (μsin28.3 + cos 28.3 )

b ) Putting the given values

= .155 x 53.1 x 9.8 x 25.3 cos28.3 / ( .155 x sin28.3 + cos 28.3 )

= 1796.76 / (.073 + .88 )

= 1885.37  J

c )

Work done by frictional force

= frictional force x displacement

=  -  μ x (mg - f sin28.3 ) x d  

= -  μ x mgd + f μsin28.3  x d

= -  μ x mgd + μsin28.3  x d x μ x mg / (μsin28.3 + cos 28.3 )

d )

Putting the values in the equation above

- .155 x 53.1 x 9.8 x 25.3 +

.155 x .474 x 25.3 x .155 x 53.1 x 9.8 /( .155 x .474 + .88)

= -2040.67 + 149.92 / .95347

= -2040.67 + 157.23

= -1883.44 J .

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