Question

a bullet of mass 5 g is fired from a rifle with a muzzle speed of 600m/s After passing through a mud wall 1 m thick the speed of the bullet drops to 400m/s. What is the work done by the mud on the bullet
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Answer 1

Given

• Initial Velocity = 600 m/s
• Final Velocity = 400 m/s
• Thickness of the wall = 1 m

To Find

• Work done by the mud wall on the bullet

Solution

☯ Work = Force × Displacement

• We have the Displacement of the body but we have to find the force First

☯ Force = Mass × Acceleration

• Here too we have only the mass of the bullet,but not the acceleration of the body

☯ v²-u² = 2as [Third Equation of Motion]

• We shall use the third equation of motion to find the acceleration of the body

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✭ According to the Question :

→ v²-u² = 2as

→ 400² - 600² = 2 × a × 1

→ 16000 - 360000 = 2a

→ -200000 = 2a

→ -200000/2 = a

→ Acceleration = -100000 m/s²

So now we can find the Force applied on the body

→ F = ma

• m = Mass = 5 g = 5/1000 = 0.005 kg
• a = Acceleration = -100000 m/s²

→ F = 0.005 × -100000

→ F = -500 N

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✭ Work Applied :

→ Work = Force × Displacement

→ Work = -500 × 1

→ Work = -500 J

• Note that here the work done is -ve as the force is against the motion of the bullet

Answer 2

Answer:

Given :-

• Mass of bullet = 5 g
• Initial velocity = 600 m/s
• Final velocity = 400 m/s
• Thickness of wall = 1 M

To Find :-

Work done

Solution :-

Firstly we will find acceleration of bullet by third Equation of motion

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

Here,

V denotes Final Velocity

U denotes initial velocity

A denotes Acceleration

S denotes Distance

$\tt \: {400}^{2} - {600}^{2} = 2(a)(1)$

$\tt \: 160000 - 360000 = 2(a)$

$\tt - 20000 = 2a$

$\tt \: a = \dfrac{ - 20000}{2}$

$\tt \: a = - 10000 \: mps$

Now,

Let's find Force applied

$\tt \: F = ma$

Here,

F denotes Force

M denotes Mass

A denotes Acceleration

5 g = 5/1000 = 0.005 kg

$\tt \: F = 0.005 \times - 10000$

$\tt \: F = - 500 \:$

Now,

Let's find work done

$\tt \: W = fs$

Here,

W denotes Work done

F denotes Force

S denotes displacement

$\tt \: W = - 500 \times 1$

$\huge \tt \pink{work \: done \: = - 500 \: J}$