Question

bullet of mass 20 g moving with speed of 80 m/s enters into a heavy wooden block and is stopped after a distance of 50 cm. The average resistive force exerted by the block on bullet is

Answer 1

Answer:

Resistive force of the bullet is -128N

Explanation:

Given,

A bullet of mass 20g moving with a velocity of 80m/s hits a wooden block and stops after a distance of 50cm is covered.

Find the average resistive force exerted by the block.

Average resistive force is given by,

$\implies \: {\rm F} = ma$

Where,

• $\boldsymbol m$ denotes the mass of the bullet.
• $\boldsymbol a$ is the acceleration.

Lets calculate the acceleration.

We have,

• $u$(intial velocity) = 80m/s
• $v$(final velocity) = 0m/s [Since it comes to rest]
• $s$(distance) = 50cm or, 0.5m [On conversion]

By using the 3rd equation of motion,

$\implies \: 2as = {v}^{2} - {u}^{2}$

Substitute the values,

$\implies \: 2(a)(0.5) = {0}^{2} - {80}^{2}$

$\implies \: 1a = - 6400$

$\implies \: a = - 6400$

Now calculate the resistive force :-

Converting mass into kg

= 20g ÷ 1000 [Since 1kg = 1000g]

= 0.020 kg

$\therefore \: \rm F = 0.020 \times ( - 6400)$

$\implies \: \rm F = - 128$

Hence, the resistive force exerted by the wooden block is -128N

[Note: -ve sign denotes the opposite force ]

__________________________

Answer 2

Answer:

Given :-

• A bullet of mass 20 g moving with speed of 80 m/s enters into a heavy wooden block and is stopped after a distance of 50 cm.

To Find :-

• What is the average resistive force exerted by the block on bullet.

Formula Used :-

$\clubsuit$ Third Equation Of Motion Formula :

$\bigstar \: \: \sf\boxed{\bold{\pink{v^2 =\: u^2 + 2as}}}\: \: \: \bigstar$

where,

• v = Final Velocity
• u = Initial Velocity
• a = Acceleration
• s = Distance Travelled or Displacement

$\clubsuit$ Force Formula :

$\bigstar \: \: \sf\boxed{\bold{\pink{F =\: m \times a}}}\: \: \: \bigstar$

where,

• F = Force
• m = Mass
• a = Acceleration

Solution :-

First, we have to find the acceleration :

Given :

• Initial Velocity (u) = 80 m/s
• Final Velocity (v) = 0 m/s
• Distance Travelled = 50 cm = 0.5 m

According to the question by using the formula we get,

$\implies \bf v^2 =\: u^2 + 2as$

$\implies \sf (0)^2 =\: (80)^2 + 2a(0.5)$

$\implies \sf (0 \times 0) =\: (80 \times 80) + a$

$\implies \sf 0 =\: 6400 + a$

$\implies \sf 0 - 6400 =\: a$

$\implies \sf - 6400 =\: a$

$\implies \sf\bold{\green{a =\: - 6400\: m/s^2}}$

[Note :- The negetive sign of acceleration denotes that the bullet is retardation in the heavy wooden block. ]

Hence, the acceleration is - 6400 m/s² .

Now, we have to find the force :

Given :

• Mass (m) = 20 g = 0.02 kg
• Acceleration (a) = - 6400 m/s²

According to the question by using the formula we get,

$\implies \bf F =\: m \times a$

$\implies \sf Force =\: Mass \times Acceleration$

$\implies \sf Force =\: 0.02 \times (- 6400)$

$\implies \sf\bold{\purple{Force =\: - 128\: N}}$

Hence, the force is - 128 N .

Now, we have to find the average resistive force exerted by the block on bullet :

Given :

• Force = - 128 N

So,

$\dashrightarrow \sf Average\: Resistive\: Force =\: - (- 128)\: N$

$\dashrightarrow \sf\bold{\red{Average\: Resistive\: Force =\: 128\: N}}$

$\therefore$ The average resistive force exerted by the block on bullet is 128 N .

▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃