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

potential energy is directly proportional to height.

more height = more potential energy.

greatest increase in height = greatest increase in potential energy

where are the points with the greatest increase in height ?

has to be c-e or d-e .