A bullet of mass 0.04 kg moving with a velocity of 90 ms-1 enters a heavy wooden block and is stopped after travelling a distance of 60 cm. Find the resistance force in Newton exerted by the block on the bullet. The first and the correct answer will be marked as the brainliest!!
Answers
Answered by
2
Answer:
Given,
m=0.04kg
u=90m/s
v=0m/s
S=60cm
From 3rd equation of motion,
2aS=v
2
−u
2
a=−
2S
u
2
a=−
2×60×10
−2
90×90
=−6750m/s
2
The resistive force, f=ma
f=−0.04×67.5
f=−270N
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Answered by
2
Answer:
The retardation 'a' of the bullet (assume constant) is given by
a = \frac{-u^2}{2s} = \frac{-90 \times 90}{2 \times 0.6} \text{ m s}^{-2} = -6750\text{ m s}^{-2}a=
2s
−u
2
=
2×0.6
−90×90
m s
−2
=−6750 m s
−2
The retarding force, by the Second Law of motion, is
= 0.04 \text{ kg} \times 6750 \text{ m s}^{-2} = 270 \text{ N}=0.04 kg×6750 m s
−2
=270 N
The actual resistive force, and therefore, retardation of the bullet may not be uniform. The answer therefore, only indicates the average resistive force.
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