# Average force exerted by object dropped onto floor [closed]

## Question :

A 2.0 kg mud ball drops from rest at a height of 15 m. If the impact between the ball and the ground lasts 0.50 s, what is the average force exerted by the ball on the ground

## working

I have this setup :

Taking gravity $g = 10$.

At $A$ potential energy $= mgh = 15(10)(2) = 300$.

Using Conservation of energy as:

$$W + PE_0 + KE_0 = PE_f + KE_f + \text{Energy(Lost)}$$

Where all energies other than $PE$ and $KE$ are zero gives

$$PE_A = KE_B$$

So that the kinetic energy when impact starts is equal to the initial potential energy, which is $300$.

From this we can find the velocity as

$$KE = \frac{1}{2}mv^2$$

Meaning

$$v = \sqrt{300} \approx 17.32$$

Using Impulse momentum theorem we have

$$I = F \Delta t = \Delta p = m (v_1 - v_0)$$

Here $v_0 = 0$ and $v_1$ has been found as $\sqrt{300}$.

From this we have

$$F \Delta t = m(\sqrt{300})$$

And

$$F = \frac{m\sqrt{300}}{\Delta t}$$

Here $\Delta t = 0.5$ then the average force exerted is $F = 4\sqrt{300} = 40 \sqrt{3} \approx 69.28$

Therefore the average force exerted is $69.28$ to 2 decimal places

## closed as off-topic by ACuriousMind♦Apr 29 '17 at 18:42

This question appears to be off-topic. The users who voted to close gave this specific reason:

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• So what is your question? – Farcher Apr 29 '17 at 18:13
• @Farcher the question is stated at the top. My working is what I'm not confident with and I don't have any solutions. – baxx Apr 29 '17 at 18:15