Answer:
Force exerted, F = 1.5 N
Explanation:
It is given that, a boxer punches a sheet of paper in midair and brings it from rest up to a speed of 30 m/s in 0.060 s.
i.e. u = 0
v = 30 m/s
Time taken, t = 0.06 s
Mass of the paper, m = 0.003 kg
We need to find the force the boxer exert on it. The force can be calculated using second law of motion as :
F = 1.5 N
So, the force the boxer exert on the paper is 1.5 N. Hence, this is the required solution.
The average force of the water droplets is the force given by the impact
per second of the droplets on the limestone floor.
Reasons:
The given parameters are;
Volume of a droplet = 10 ml = 1 × 10⁻⁵ m³
Height from which the water falls, <em>h </em>= 5 meters
Rate at which the water falls = 10 per minute
Required:
The average force exerted on the floor by the water droplets.
Solution:
According to Newton's Second Law of motion, we have;
Force = Rate of change of momentum
Momentum = Mass × Velocity
Mass of a droplet of water = Volume × Density
Density of water = 997 kg/m³
Mass of a droplet = 1 × 10⁻⁵ m³ × 997 kg/m³ = 0.00997 kg
The velocity just before the droplet reaches the ground, v = √(2·g·h)
Where;
g = Acceleration due to gravity ≈ 9.81 m/s²
Which gives;
v = √(2 × 9.81 m/s² × 5 m) ≈ 9.905 m/s
The rate of change in momentum per minute = 1
Therefore;
ΔMomentum = Mass × ΔVelocity
Considering the 10 drops per minute, we have;
ΔMomentum = 10 × 0.0097 kg × 9.905 m/s = 0.960785 kg·m/s
ΔTime = 1 minute = 60 seconds
Therefore;
Learn more about Newton's Second Law of motion and force exerted water here: