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### JEE Main 2016 (Online) 10th April Morning Slot

Velocity-time graph for a body of mass 10 kg is shown in figure. Work-done on the body in first two seconds of the motion is :

A
12000 J
B
$-$ 12000 J
C
$-$ 4500 J
D
$-$ 9300 J

## Explanation

Here u = 50 m/s , what t = 0

$\alpha$  =  ${{\Delta v} \over {\Delta t}}$  =  ${{50 - 0} \over {0 - 10}}$  =  $-$5 m/s2

Speed of the body at t = 2 s

v   =   u + at

=  50 + ($-$ 5) $\times$ 2

=  40 m/s

From work energy theorem,

$\Delta$w  =  ${1 \over 2}m{v^2} - {1 \over 2}m{u^2}$

=  ${1 \over 2}$ m(v2 $-$u2)

= ${1 \over 2}$ $\times$ 10 $\times$ (402 $-$ 502)

=  5 $\times$ ($-$10)(90)

=  $-$ 4500 J
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### JEE Main 2017 (Offline)

A body of mass m = 10–2 kg is moving in a medium and experiences a frictional force F = –kv2. Its initial speed is v0 = 10 ms–1. If, after 10 s, its energy is ${1 \over 8}mv_0^2$, the value of k will be:
A
10-1 kg m-1 s-1
B
10-3 kg m-1
C
10-3 kg s-1
D
10-4 kg m-1

## Explanation

According to the question, final kinetic energy = ${1 \over 8}mv_0^2$

Let final speed of the body = Vf

So final kinetic energy = ${1 \over 2}mv_f^2$

According to question,

${1 \over 2}mv_f^2$ = ${1 \over 8}mv_0^2$

$\Rightarrow {v_f} = {{{v_0}} \over 2}$ = ${{10} \over 2}$ = 5 m/s

Given that, F = –kv2

$\Rightarrow$ $m\left( {{{dv} \over {dt}}} \right)$$= - k{v^2} \Rightarrow {10^{ - 2}}\left( {{{dv} \over {dt}}} \right) = - k{v^2} \Rightarrow \int\limits_{10}^5 {{{dv} \over {{v^2}}}} = - 100k\int\limits_0^{10} {dt} \Rightarrow {1 \over 5} - {1 \over {10}} = 100k \times 10 \Rightarrow k = {10^{ - 4}}kg\,{m^{ - 1}} 3 MCQ (Single Correct Answer) ### JEE Main 2017 (Offline) A time dependent force F = 6t acts on a particle of mass 1 kg. If the particle starts from rest, the work done by the force during the first 1 sec. will be: A 18 J B 4.5 J C 22 J D 9 J ## Explanation Given that, F = 6t We know, F = ma = m{{dv} \over {dt}} \therefore m{{dv} \over {dt}} = 6t \Rightarrow 1.{{dv} \over {dt}} = 6t [as m = 1] \Rightarrow \int\limits_0^v {dv} = \int {6t} dt \Rightarrow v = 6\left[ {{{{t^2}} \over 2}} \right]_0^1 \Rightarrow v = {6 \over 2} = 3 m/s [ as given t = 1 sec ] Work done by the body during the first 1 form work-energy theorem, W = \Delta K.E = {1 \over 2}m\left( {{V^2} - {v^2}} \right) = {1 \over 2}.1.\left( {{3^2} - {0^2}} \right) = 4.5 J 4 MCQ (Single Correct Answer) ### JEE Main 2017 (Online) 8th April Morning Slot An object is dropped from a height h from the ground. Every time it hits the ground it looses 50% of its kinetic energy. The total distance covered as t \to \infty is : A 3h B \infty C {5 \over 3}h D {8 \over 3}h ## Explanation Let, Kinetic energy (k) = {1 \over 2} m \upsilon 2 before it hit the ground. After hitting the ground kinetic energy (k') = {1 \over 2} m \upsilon$$_1^2$

$\therefore\,\,\,$According to the question,

${1 \over 2}$ m$\upsilon $$_1^2 = {1 \over 2} \times {1 \over 2} m\upsilon 2 \Rightarrow$$\,\,\,$ $\upsilon$1 = ${v \over {\sqrt 2 }}$

After hitting the ground the object will bounce

h' = ${{v_1^2} \over {2g}}$ = ${{{v^2}} \over {4g}}$ = ${h \over 2}$ [ as    h = ${{{v^2}} \over {2g}}$ ]

Total distance travelled from the time it first hits the ground to the next time it hits the ground is = ${h \over 2}$ + ${h \over 2}$ = h

So, this will create a infinite geometric progression with the common ration ${1 \over 2}$.

$\therefore\,\,\,$ Total distance covered

= h (distance travelled by the obhect when first dropped, before it hits the ground)
+ (h + ${h \over 2}$ + ${h \over 4}$ + . . . . . . . .$\propto$)

= h + ${h \over {1 - {1 \over 2}}}$

= h + 2h

= 3h