Question

the velocity of a body moving in a straight line is increased by applying a constant force F, for some distance in the direction of the motion. Prove that the increase in the kinetic energy of the body id equal to the work done by the force on the body

Answer 1

Answer:

Explanation;∵ a force is applied

∴ velocity will increase

let the new velocity = v

initial velocity = u

mass of the object = m

acceleration = a

distance covered = d

∴ by 3rd eq. of motion

v² = u² + 2ad ........................ eq(i)

increase in the kinetic energy,

new KE - initial KE

= mv²/2 - mu²/2

= m(u² + 2ad)/2 - mu²/2 ........................................ {from eq.(i)}

=mu²/2 + mad - mu²/2

= mad

=Fd .............................................. (∵F=ma)

= work done by the force on the body .................................(∵work = Fd)

∴increase in the kinetic energy= work done by the force on the body

Answer 2

Answer:

We know that $v^{2}$ = $u^{2}$ + $2as$

And Work done $F=s$ for a constant force F.

Also we know that $a=F/m$

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

This gives

$s=(v^{2} -u^{2} )/2a$

$F=ma$

We can write work done (W) by this force F as

$W=ma (v^{2} -u^{2} )/2a$$=1/2 mv^{2} -1/2mu^{2} =KEf-KEi$

Explanation: