1

### JEE Main 2018 (Online) 16th April Morning Slot

A body of mass m starts moving from rest along x-axis so that its velocity varies as $\upsilon = a\sqrt s$ where a is a constant and s is the distance covered by the body. The total work done by all the forces acting on the body in the first t seconds after the start of the motion is :
A
${1 \over 8}\,$ m a4 t2
B
8 m a4 t2
C
4 m a4 t2
D
${1 \over 4}\,$ m a4 t2

## Explanation

Given,

$\upsilon$ = a $\sqrt s$

$\Rightarrow $$\,\,\, {{ds} \over {dt}} = a\sqrt s \Rightarrow$$\,\,\,$ $\int\limits_0^t {{{ds} \over {\sqrt s }}} = \int\limits_0^z {a\,dt}$

$\Rightarrow $$\,\,\, 2\sqrt s = at \Rightarrow$$\,\,\,$ s = ${{{a^2}{t^2}} \over 4}$

= ${1 \over 2}.{{{a^2}} \over 2}.{t^2}$

$\therefore\,\,\,\,$ acceleration = ${{{a^2}} \over 2}$

$\therefore\,\,\,$ Force (F) = m $\times$ ${{{a^2}} \over 2}$

$\therefore\,\,\,\,$ Work done = F. S

= ${{m{a^2}} \over 2} \times {{{a^2}{t^2}} \over 4}$

= ${{m{a^4}{t^2}} \over 8}$
2

### JEE Main 2019 (Online) 9th January Morning Slot

A block of mass m, lying on a smooth horizontal surface, is attached to a sring (of negligible mass) of spring constant k. The other end of the spring is fixed, as shown in the figure. The block is initially at rest in its equilibrium position. If now the block is pulled with a constant force F, the maximum speed of the block is : A
${{2F} \over {\sqrt {mk} }}$
B
${F \over {\pi \sqrt {mk} }}$
C
${{\pi F} \over {\sqrt {mk} }}$
D
${F \over {\sqrt {mk} }}$

## Explanation When block is pulled x distance with F force, then it is distance with F force, then it is at maximum position

$\therefore$   F = Kx

$\Rightarrow$   x = ${F \over K}$

Applying work energy theorem, work done by all the forces = change in Kinetic energy.

$\Rightarrow$  WForce + WSpring = ${1 \over 2}m{v^2} - 0$

$\Rightarrow$   Fx $-$ ${1 \over 2}$Kx2 = ${1 \over 2}m{v^2}$

$\Rightarrow$   $F$$\left( {{F \over K}} \right) - {1 \over 2}K{\left( {{F \over K}} \right)^2}$ = ${1 \over 2}m{v^2}$

$\Rightarrow$   ${{{F^2}} \over K} - {{F{}^2} \over {2K}}$ = ${1 \over 2}m{v^2}$

$\Rightarrow$   ${{F{}^2} \over {2K}}$ = ${1 \over 2}m{v^2}$

$\Rightarrow$   $v$ = ${F \over {\sqrt {mK} }}$
3

### JEE Main 2019 (Online) 9th January Morning Slot

Three blocks A, B and C are lying on a smooth horizontal surface, as shown in the figure. A and B have equal masses, m while C has mass M, Block A is given an initial speed $\upsilon$ towards B due to which it collides with B perfectly inelastically. The combined mass collides with C, also perfectly inelastically ${5 \over 6}$th of the initial kinetic energy is lost in whole process. What is value of M/m ? A
5
B
2
C
4
D
3

## Explanation

As in elastic or in elastic clollision, momentum is conserved.

$\therefore$   Pi = Pf

Pi = Initial momentum

Pf = Final Momentum

mv = (2m + m) Vf

$\Rightarrow$   Vf = ${{mv} \over {2m + M}}$

Here due to collision ${5 \over 6}$th of kinetic energy is lost.

$\therefore$    Remaining kinetic energy,

Kf = ${1 \over 6}$ Ki

$\Rightarrow$    ${1 \over 2}$(2m + M) $\times$ ${{{{\left( {mv} \right)}^2}} \over {{{\left( {2m + M} \right)}^2}}}$ = ${1 \over 6} \times {1 \over 2}m{v^2}$

$\Rightarrow$   ${{{m^2}{v^2}} \over {2m + M}}$ = ${1 \over 6}m{v^2}$

$\Rightarrow$   ${m \over {2m + M}}$ = ${1 \over 6}$

$\Rightarrow$   6m = 2m + M

$\Rightarrow$   M = 4m

$\Rightarrow$    ${M \over m}$ = 4
4

### JEE Main 2019 (Online) 9th January Evening Slot

A force acts on a 2 kg object so that its position is given as a function of time as x = 3t2 + 5. What is the work done by this force in first 5 seconds ?
A
850 J
B
950 J
C
875 J
D
900 J

## Explanation

Displacement,

x = 3t2 + 5

$\therefore$  v = ${{dx} \over {dt}} = 6t$

At t = 0,   velocity = 6 $\times$ 0 = 0

at t = 5, velocity = 5 $\times$ 6 = 30 m/s

we know from work energy theorem,

Work (W) = change in kinetic energy ($\Delta$K)

= ${1 \over 2}mv_F^2 - {1 \over 2}mv_i^2$

= ${1 \over 2}$ $\times$ 2 $\times$ (30)2 $-$ 0

= 900 J