2.

A sled with a boy on it has a combined mass of m = 75 kg. As they slide across a horizontal surface of an ice-

covered lake, they come to rest in a distance of s = 25 m. If the coefficient of friction between the sled and the ice

is u = 0.25, calculate the amount of work done on the sled by friction. Keep in mind - negative Work decreases

Kinetic Energy whereas Positive Work increases it. So, don't forget a negative (-) sign in front of your answer.

The work done on the sled by friction, W = - 4593.75 J

Explanation:

Given data,

The combined mass of sled and the boy, m = 75 kg

The displacement of the boy, S = 25 m

The coefficient of the friction, u = 0.25

The frictional force acting on the boy,

F = u η

Where,

η - is the normal force acting on the boy (mg)

Substituting the values,

F = 0.25 x 75 x 9.8

= 183.75 N

Since the direction of the frictional force is against the direction of motion

F = - 183.75 N

The work done on the sled by friction,

W = F x S

= - 183.75 x 25

= - 4593.75 J

Hence, the work done on the sled by friction, W = - 4593.75 J