How Is Force Calculated When an Object Hits the Ground?

[cd19]

An object of mass 4 kg is released from rest at a height of 10 m. After striking the ground it take 20ms before it is brought to rest. Assuming that the object exerts a constant force F on the ground during the 20ms interval then what is the magnitude of F?

Having trouble understanding the question, would appreciate if someone could put it into perspective.

 

[LowlyPion]

An object of mass 4 kg is released from rest at a height of 10 m. After striking the ground it take 20ms before it is brought to rest. Assuming that the object exerts a constant force F on the ground during the 20ms interval then what is the magnitude of F?

Having trouble understanding the question, would appreciate if someone could put it into perspective.

Welcome to PF.

Consider the speed that the object has acquired by falling 10m.

F = m*a = m*Δv/Δt = Δmv / Δt

 

[cd19]

I have found the speed acquired by falling 10m to be 14 ms (m*g*h=1/2*m*v^2). but I still can't comprehend what to do next.

 

[Cyosis]

Remember what force is. F=ma, with a the acceleration as you know. Once the object hits the ground it goes from 14m/s to 0s in 20ms. This means it is decelerating. It has mass and it is decelerating that sounds like the two essential components to calculate the force. The basic question is what is its deceleration?

 

[LowlyPion]

I have found the speed acquired by falling 10m to be 14 ms (m*g*h=1/2*m*v^2). but I still can't comprehend what to do next.

F = m*a = m*Δv/Δt = Δmv / Δt

you know m,
you know v,
you know Δt ...

 

[cd19]

ah i see, I've been reading the question wrong, I'll put it down to tiredness. I kept thinking 20 ms was 20 ms^-1. how stupid. I found the force to be 2800N, which i feel is correct. thanks very much for the quick reply's.

 

[LowlyPion]

I found the force to be 2800N, which i feel is correct. thanks very much for the quick reply's.

Looks right.

Good Luck.