Question

A bullet of mass 30g strikes a target with velocity of 90 m/s and is brought to rest after penetrating 15cm into it . Calculate work done by the bullet​

Answer 1

The work done by the bullet is W = - 121.5 J

Explanation:

Mass m =m 30 g 0.03 Kg

• u = 90 m/s
• v = 0 m/s
• s = 15 cm = 0.15 m

By work energy theorem:

Net work done = Change in K.E

W = 1/2 mv^2 - 1/2 mu^2

W = 1/2 m (v^2 - u^2)

W = 1/2 x 0.03 ( 0^2 - 90^2)

W = - 1/2 x 0.03 x 8100

W = - 121.5 J

Thus the work done by the bullet is W = - 121.5 J

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Answer 2

Answer:

• Work done (W) by the Bullet is - 121.5 Joules.

Given:

1. Mass of Bullet (M) = 30 grams
2. Initial velocity (u) = 90 m/s
3. Penetration Distance (S) = 15 cm

Explanation:

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By Work - Energy Theorem,

⟼ Work Done = Change in Kinetic Energy.

⟼ W = Δ K.E

⟼ W = 1 / 2 M v² - 1 / 2 M u²

Substituting the Values,

⟼ W = 1 / 2 x 30 grams x v² - 1 / 2 x 30 grams x (u)²

∵ [1 gram = 10⁻³ Kg ⟼30 grams = 30 x 10⁻³ Kg]

⟼ W = 1 / 2 x 30 x 10⁻³ Kg x v² - 1 / 2 x 30 x 10⁻³ Kg x (u)²

Here,

• Final Velocity = 0 m/s ( As body comes to rest)
• Initial velocity = 90 m/s

Now,

⟼ W = 1 / 2 x 30 x 10⁻³ x 0 - 1 / 2 x 30 x 10⁻³ x (90)²

⟼ W = - 1 / 2 x 30 x 10⁻³ x 8100

⟼ W = - 1 / 2 x 3 x 10⁻² x 8100

⟼ W = - 1 / 2 x 24300 x 10⁻²

⟼ W = - 1 / 2 x 243

⟼ W = - 121.5

⟼ W = - 121.5 Joules.

∴ Work done (W) by the Bullet is - 121.5 Joules.

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