Question

A boxer punches a sheet of paper in mid air, and brings it from rest up to a speed of 25 m/s in 0.05 s. if the mass of the paper is 0.005 kg, what force does the boxer exert on it

Answer 1

Answer:

F = 2.5 N

Explanation:

Because the sheet of paper moves with a uniformly accelerated movement, we apply the following formulas:

vf= v₀+at Formula (1)

Where:  

t : time in seconds (s)

v₀: initial speed in m/s  

vf: final speed in m/s  

a: acceleration in m/s²

We apply Newton's second law:

F=m*a Formula (2)

F ; Force (N)

m: mass  (kg)

a: acceleration  (m/s² )

Problem development

We calculate the acceleration using the formula (1):

vf= v₀+at   v₀ = 0, vf = 25 m/s ,  t= 0.05 s

vf-v₀ = at

a =  (vf -v₀ ) /(t)

a = (25 -0)/(0.05)

a = 500 m/s²

We calculater the force exerted by the boxer on the sheet of paper (F) using formula (2):

F= m*a

F= 0.005 kg*500 m/s²

F = 2.5 N

Answer 2

Ok, so you've got to figure out a force F and you have the speed in which the boxer punches on determinate time and the mass of the sheet of paper.

So based on the formula that says that the Force is equal to the mass multiplied by the acceleration => F=ma.
You look at it and see that you only have mass which is measured on KG so there is no problem.
then you have the acceleration which is measured on meters and is defined by: a = Δv/Δt 
So now you can replace the velocity and the time you have there
⇒ a 25m/s / 0.05s 
you have computing that ⇒ 50m because the seconds were cancelled out.
and then you plug the meters into the force equation.
F=(0.005kg)(50)
F=0.25N
so the boxer will have a force of 0.25 Newton's.