1) A 0.25 kg soccer ball is rolling at 6.0m/s toward a player. The player kicks the ball back in the oppoiste direction and gives it a velocity of -14 m/s. What is the average force between the players force during the interaction between the players foot and the ball if the interaction lasts 2.0 X 10^-2 seconds.
Rather than tease you with the answer like one might a dog with a packet of dog biscuits, I think I'll just give you the answer.
You already know that force is equal to mass multiplied by an acceleration, which is a change in velocity over time.
F = ma = m(Δv/Δt)
Writing this differently,
F = (mΔv)/Δt
You may notice that the numerator can also be written Δmv, a change in momentum. This is another definition of a force: a change of momentum over time. This will give you your answer.
A change in momentum is
Δp = mv2 - mv1
Calling the final and initial velocities v2 and v1 respectively.
Since m is common to both terms,
Δp = m(v2 - v1)
Now, the change in velocity is the final velocity minus the initial velocity. Calling the direction in which the ball is kicked by the player the positive direction, the change in velocity is
v2 - v1 = 14 - (-6) = 14 + 6 = 20 ms-1
You need to pay attention to the signs. Plugging this answer into the equation worked out above:
F = mΔv/Δt = m(20)/(1/50) = (50 * 20)m = 1000m
A precise numerical answer is 1000 * 0.25 = 250 N.
2)A person weighing 490 N stands on a scale in an elevator.
- The elevator slows down at -2.2 m/s as it reaches the desired floor, what does the scale read?
Now F = ma, so another way of thinking about acceleration is the force acting per unit mass, a = F/m. Now your weight is equal to
W = mg
Where g is the acceleration due to gravity at the Earth's surface, this is roughly 9.8ms-2.
Now, the force acting on the scales when the lift is moving at constant velocity is equal to the person's weight. Gravity is pulling the man down with a force of 9.8 Newtons per kilogram. But there's another force acting on the man when the lift is decelerating.
Let's assume that the lift is traveling upward and the upward is the positive direction. The lift will slow down as it reaches the floor and as a result, the reading on the scale will decrease.
Call the force on the scale F, then
F = W + ma
Where W is the man's weight, m his mass and a the acceleration of the lift.
F = W + m(-2.2) = W - 2.2m
m is W/g, so
F = W - 2.2(W/g) = g(W/g) - 2.2(W/g) = (W/g)(9.8 - 2.2) = (W/g)(7.7)
This is equal to (420 * 7.7)/9.8 = 385 N