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Work Energy Theorem

Same force ac...

Question

Same force acts on two bodies of masses $m_{1}$ and $m_{2}$ ( $m_{2}$ > $m_{1}$ ). After travelling the same distance,

Kinetic energy of

m_{2}

is greater than kinetic energy of

m_{1}

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Kinetic energy of

m_{1}

is greater than kinetic energy of

m_{2}

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Ratio of kinetic energies of those two is equal to ratio of their momentum

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Kinetic energy of

m_{1}

is equal to kinetic energy of

m_{2}

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Solution

The correct option is D Kinetic energy of $m_{1}$ is equal to kinetic energy of $m_{2}$
$s = \frac{1}{2} a t^{2}$
Now, $K . E_{1} = \frac{1}{2} m_{1} (a t)^{2}$ $= \frac{1}{2} \frac{m_{1}}{a_{1}} s a_{1} = F . s$
$K . E_{2} = \frac{1}{2} m g (a_{2} t)^{2}$ $= F . s$
i.e $K . E_{1} = K E_{2}$

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Same force acts on two bodies of masses m1 and m2 (m2> m1). After travelling the same distance,

The correct option is D Kinetic energy of m1 is equal to kinetic energy of m2s=12at2Now, K.E1=12m1(at)2=12m1a1sa1=F.sK.E2=12mg(a2t)2=F.si.e K.E1=KE2

Same force acts on two bodies of masses $m_{1}$ and $m_{2}$ ( $m_{2}$ > $m_{1}$ ). After travelling the same distance,

The correct option is D Kinetic energy of $m_{1}$ is equal to kinetic energy of $m_{2}$
$s = \frac{1}{2} a t^{2}$
Now, $K . E_{1} = \frac{1}{2} m_{1} (a t)^{2}$ $= \frac{1}{2} \frac{m_{1}}{a_{1}} s a_{1} = F . s$
$K . E_{2} = \frac{1}{2} m g (a_{2} t)^{2}$ $= F . s$
i.e $K . E_{1} = K E_{2}$