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A pile driver of $150 kg$ falls from a height of $5 m$ above the pile and comes to rest in $0.5 s$ . What will be the average force exerted on the pile?

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- 5400 N

4600 N

- 3000 N

None of these

answer is C.

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Detailed Solution

Given that,
The average force exerted on the pile is

- 3000 N

.
Mass,

m = 150 kg

, Height,

h = 5 m

, Initial velocity of the pile driver,

u = 0 m / s

Let the velocity of the pile driver when it touches the pile is

= v

Time

t = 0.5 s

We know that

v^{2} = u^{2} + 2 as

Where

v

is the final velocity,

u

is the initial velocity,

a

is the acceleration and

s

is the displacement.
Now in place of

a

and

s

we can take

g

and

h

respectively.

v^{2} = u^{2} + 2 gh

…(1)
Where

g

is the gravitational acceleration and

h

denotes the height.
Put the given values in the above equation,

v^{2} = 0 + 2 \times 10 \times 5

\Rightarrow v^{2} = 100

\Rightarrow v = \sqrt{100}

\Rightarrow v = 10 m / s

Now

v = 10 m / s

is the initial velocity of the pile,
Final velocity of the pile is,

v_{1} = 0

[as it stops after

0.5

seconds] Time,

t_{1} = 0.5 seconds

Now we know that net unbalanced force equals the rate of change of momentum and it can be formulated as follows,

F = \frac{m v_{1} – mv}{t}

\Rightarrow F = \frac{0 - 150 kg \times 10 m / s}{0.5 s}

\Rightarrow F = - 3000 kg m / s^{2}

A negative sign indicates that the direction of applied force is opposite.

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