JEE Main Work, Energy and Power PYQ

16. (JEE Main 2019 (Online) 10th January Morning Slot )

A block of mass m is kept on a platform which starts from rest with constant acceleration g/2 upward, as shown in figure. Work done by normal reaction on block in time is -

JEE Main 2019 (Online) 10th January Morning Slot Physics - Work Power & Energy Question 73 English

A. $\frac{m g^{2} t^{2}}{8}$

B. $\frac{3 m g^{2} t^{2}}{8}$

C. $-$ $\frac{m g^{2} t^{2}}{8}$

D. 0

Correct Option is (B)

N $-$ mg = $\frac{m g}{2}$ $\Rightarrow$ N = $\frac{3 m g}{2}$

The distance travelled by the system in time t is

S = ut + $\frac{1}{2} a t^{2} = 0 + \frac{1}{2} (\frac{g}{2}) t^{2} = \frac{1}{2} \frac{g}{2} t^{2}$

Now, work done

W = N.S = $(\frac{3}{2} m g) (\frac{1}{2} \frac{g}{2} t^{2})$

$\Rightarrow$ W = $\frac{3 m g^{2} t^{2}}{8}$

17. (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 $υ = a \sqrt{s}$ 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. $\frac{1}{8}$ m a⁴ t²

B. 8 m a⁴ t²

C. 4 m a⁴ t²

D. $\frac{1}{4}$ m a⁴ t²

Correct Option is (A)

Given,

$υ$ = a $\sqrt{s}$

$\Rightarrow$ $\frac{d s}{d t} = a \sqrt{s}$

$\Rightarrow$ $\int_{0}^{t} \frac{d s}{\sqrt{s}} = \int_{0}^{z} a d t$

$\Rightarrow$ 2 $\sqrt{s}$ = at

$\Rightarrow$ s = $\frac{a^{2} t^{2}}{4}$

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

$∴$ acceleration = $\frac{a^{2}}{2}$

$∴$ Force (F) = m $\times$ $\frac{a^{2}}{2}$

$∴$ Work done = F. S

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

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

18. (JEE Main 2016 (Offline) )

A person trying to lose weight by burning fat lifts a mass of $10$ $k g$ upto a height of $1$ $m$ $1000$ times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up? Fat supplies $3.8 \times 10^{7} J$ of energy per $k g$ which is converted to mechanical energy with a $20 %$ efficiency rate. Take $g = 9.8 m s^{- 2}$ :

A. $9.89 \times 10^{- 3} k g$

B. $12.89 \times 10^{- 3} k g$

C. $2.45 \times 10^{- 3} k g$

D. $6.45 \times 10^{- 3} k g$

Correct Option is (B)

Assume the amount of fat is used = x kg

So total Mechanical energy available through fat

= $x \times 3.8 \times 10^{7} \times \frac{20}{100}$

And work done through lifting up

= 10 $\times$ 9.8 $\times$ 1000 = 98000 J

$\Rightarrow$ $x \times 3.8 \times 10^{7} \times \frac{20}{100}$ = 98000

$\Rightarrow$ $x$ = 12.89 $\times$ 10^-3 kg

19. (JEE Main 2014 (Offline) )

When a rubber-band is stretched by a distance $x$ , it exerts restoring force of magnitude $F = a x + b x^{2}$ where $a$ and $b$ are constants. The work done in stretching the unstretched rubber-band by $L$ is :

A. $a L^{2} + b L^{3}$

B. $\frac{1}{2} (a L^{2} + b L^{3})$

C. $\frac{a L^{2}}{2} + \frac{b L^{3}}{3}$

D. $\frac{1}{2} (\frac{a L^{2}}{2} + \frac{b L^{3}}{3})$

Correct Option is (C)

Given Restoring force, F = ax + bx²

Work done in stretching the rubber-band by a distance $d x$ is

$d W = F d x = (a x + b x^{2}) d x$

Intergrating both sides,

$W = \int_{0}^{L} a x d x + \int_{0}^{L} b x^{2} d x$

= ${[a \frac{x^{2}}{2} + b \frac{x^{3}}{3}]}_{0}^{L}$

= $\frac{a L^{2}}{2} + \frac{b L^{3}}{3}$

20. (AIEEE 2012 )

This question has Statement $1$ and Statement $2.$ Of the four choices given after the Statements, choose the one that best describes the two Statements.

If two springs $S_{1}$ and $S_{2}$ of force constants $k_{1}$ and $k_{2}$ , respectively, are stretched by the same force, it is found that more work is done on spring $S_{1}$ than on spring $S_{2}$ .

STATEMENT 1: If stretched by the same amount work done on $S_{1}$ , Work done on $S_{1}$ is more than $S_{2}$
STATEMENT 2: $k_{1} < k_{2}$

A. Statement 1 is false, Statement 2 is true

B. Statement 1 is true, Statement 2 is false

C. Statement 1 is true, Statement 2 is true, Statement 2 is the correct explanation for Statement 1

D. Statement 1 is true, Statement 2 is true, Statement 2 is not the correct explanation for Statement 1

Correct Option is (A)

We know force (F) = kx

$W = \frac{1}{2} k x^{2}$

$W =$ $\frac{{(k x)}^{2}}{2 k}$

$∴$ $W = \frac{F^{2}}{2 k}$ [ as $F = k x$ ]

When force is same then,

$W \propto \frac{1}{k}$

Given that, $W_{1} > W_{2}$

$∴$ $k_{1} < k_{2}$

Statement-2 is true.

For the same extension, x₁ = x₂ = x

Work done on spring S₁ is W₁ = $\frac{1}{2} k_{1} x_{1}^{2} = \frac{1}{2} k_{1} x^{2}$

Work done on spring S₂ is W₂ = $\frac{1}{2} k_{2} x_{2}^{2} = \frac{1}{2} k_{2} x^{2}$

$∴$ $\frac{W_{1}}{W_{2}} = \frac{k_{1}}{k_{2}}$

As $k_{1} < k_{2}$ then $W_{1} < W_{2}$

So, Statement-1 is false.