A68.5-kg skater moving initially at 2.40 m> s on rough horizontal ice comes to rest uniformly in 3.52 s due to friction from the ice. what force does friction exert on the skater?

To determine the friction force exerted in the skater, we use Newton's Second law of motion which relates force, acceleration and the mass. It is expressed as follows: 

F = m (a) 

where
m = mass of the body = 68.5 kg 

The acceleration, a,  of the moving body  is calculated by dividing the initial velocity with the total time. Thus,

a = 2.40 m/s / 3.52 s = 0.682 m/s^2 

Substituting the values to the equation of force, we will have

F = m (a) = 68.5 kg ( 0.682 m/s^2 ) = 46.7 N 

Thus, the friction force that is being exerted by the ice  to the skater ould be 46.7 N.

W= fd
where f is an applied force and d is the distance over which it is applied. when a parachuting skydiver is at constant velocity, he is experiencing "terminal velocity", the point at which his acceleration down is canceled out by air resistance. that means that the resisting force of the air is equal to his weight, but in the opposite direction. with this, we can solve for the work done by the force.
w = fd
w = (92.0)(9.81 m/s^2) (325)
w= 2.93 x 10^5 j
the value for work in this case is positive because it is pushing up on the skydiver, who has a downward force of gravity.
therefore the answer is 2.93 x 10^5 j

All are great precautions but you can still lift lighter and different muscles in between hard workdays. this keeps other muscles not usually used in shape.