A bullet of mass $100 g$ has attained a velocity of $1080 k m h^{- 1}$ towards the target. What would its kinetic energy be at that moment?

4500 J

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4240 J

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4300 J

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4750 J

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Solution

The correct option is A $4500 J$
Given, mass of the body, $m = 100 g = 0.1 k g$ and velocity of the body, $v = 1080 k m h^{- 1} = \frac{1080 \times 1000}{60 \times 60}$
$= 300 m s^{- 1}$
Kinetic energy, $K . E . = \frac{1}{2} m v^{2}$
Substituting the values in the expression of K.E, we get:
$K . E = \frac{1}{2} \times 0.1 \times 300^{2} = 4500 J$

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A bullet of mass 100 g has attained a velocity of 1080 km h−1 towards the target. What would its kinetic energy be at that moment?

The correct option is A 4500 JGiven, mass of the body, m=100 g=0.1 kg and velocity of the body, v=1080 km h−1=1080×100060×60 =300 ms−1 Kinetic energy, K.E.=12mv2 Substituting the values in the expression of K.E, we get: K.E =12×0.1×3002 = 4500 J

A bullet of mass $100 g$ has attained a velocity of $1080 k m h^{- 1}$ towards the target. What would its kinetic energy be at that moment?