Question

A bullet of mass 40g moving with a speed of 90m/s enters a heavy wooden block and it is stop after a distance of 60cm.The average resistive force exited by the block on the bullet.

Answer 1

Answer:

-270 newtons

Explanation:

Mass of the bullet is 40 g or 0.04 kg. The initial velocity ( u ) is 90 m/s and displacement ( distance ) is 60 cm or 0.6 m. The final velocity ( v ) is 0 m/s ( the bullet comes to rest after hitting the block ).

We are having,

$m = 0.04 \ kg\\u = 90 \ m/s\\s = 0.6 \ m$

Using,

$v^2 = u^2 + 2as$

We can find the acceleration applied on the bullet by the block.

$0 = 8100 + 2a( 0.6 )\\-8100 = 1.2a\\a = \frac{-8100}{1.2}\\a = -6750 \ m/s^2$

The acceleration is negative meaning that the velocity was decreasing which is true in our case.

We know,

$F = ma$

We are having the value of mass ( m ) and acceleration ( a ),

$F = 0.04 \times -6750\\F = -270 \ N$

Hence, a force of -270 newtons was applied by the block on the bullet.