Final answer:
The average force exerted by the floor on the 20.0 kg ball of wet clay that fell for 4.0 seconds and came to rest in 5.0 ms is calculated using the impulse-momentum theorem. The force is found to be 156,960 N directed upwards.
Step-by-step explanation:
The question is asking to calculate the average force exerted on a 20.0 kg ball of wet clay that falls from a height and comes to rest upon hitting the floor. The question provides the amount of time the ball falls (4.0 s) and the time it takes to come to rest (5.0 ms), and we are tasked with determining the force during this very brief stopping period.
To solve this, we first need to determine the velocity of the clay ball just before it hits the ground. We can do this by using the kinematic equation:
final velocity (v) = initial velocity (u) + acceleration (a) × time (t)
Assuming the ball starts from rest and only the acceleration due to gravity (approximately 9.81 m/s²) is acting on it:
v = 0 + (9.81 m/s²) × (4.0 s) = 39.24 m/s
After finding the velocity, we calculate the momentum change (Δp) during the impact:
Δp = mass (m) × velocity change (Δv) = 20.0 kg × 39.24 m/s = 784.8 kg·m/s
Knowing the time it takes to stop (5.0 ms or 0.005 s), we apply the impulse-momentum theorem, which states that impulse equals the change in momentum:
Impulse (J) = Force (F) × Time (Δt) = Δp
Hence, the average force exerted by the floor on the clay is:
F = Δp / Δt = 784.8 kg·m/s / 0.005 s = 156,960 N
Therefore, the floor exerted an average force of 156,960 N on the clay to bring it to rest in 5.0 ms. The direction of this force is upwards, opposing the motion of the falling clay.