Question

A bullet of mass 0.04kg moving with a speed of 90m/sec enters a heavy wooden block and is stopped after a distance of 60cm. what is the average resistive force exerted by the block on the bullet?

Answer 1

Here, mass= 0.04 kg
U= 90meter per sec
V= 0
S= 0.6 m
From, 3rd equation of motion
V2 - u2 = 2as
Acceleration = - 7750
From, 2nd law of motion
F =ma
F = - 310 N
From, Newton's 3rd law,
F1 = - F2
Therefore, average resitive foce exerted by the block on the bullet= 310 N

Answer 2

Given :

mass = m= 0.04kg

initial velocity = u=90m/s

final velocity = v= 0m/s

displacement = s= 60cm =0.6m

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Need to Find:

The Force acting on body

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Solution

:

By 3rd equation of motion we know,

${v}^{2} - {u}^{2} = 2as$

$\implies \: {0}^{2} - {90}^{2} = 2a \times60 \\ </p><p></p><p>\implies \: - 8100 = 0.6 \times 2 \times a \\ </p><p></p><p></p><p>\implies \: a = \dfrac{ - 81000}{2 \times 6} \\ </p><p> </p><p> </p><p></p><p>\purple{\bold{\boxed{ \implies a= 6750 m/ s^{2}}}} </p><p></p><p> </p><p> </p><p>$

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We know,

• F= m× a

• ⟹ F= 0.04 × (-6750) N

$⟹ \orange{\bold{\boxed{\large{Force= (-270)N}}}}$

• Force=(−270)N

• The Force of 270N is acting in opposite to the direction of motion.