JEE Main Work, Energy and Power PYQ

16. (JEE Main 2016 (Online) 9th April Morning Slot )

A car of weight W is on an inclined road that rises by 100 m over a distance of 1 km and applies a constant frictional force $\frac{W}{20}$ on the car. While moving uphill on the road at a speed of 10 ms⁻¹, the car needs power P. If it needs power $\frac{p}{2}$ while moving downhill at speed v then value of $υ$ is :

A. 20 ms^{$-$ 1}

B. 15 ms^{$-$ 1}

C. 10 ms^{$-$ 1}

D. 5 ms^{$-$ 1}

Correct Option is (B)

Here, tan $θ$ = $\frac{100}{1000} = \frac{1}{10}$

$∴$    sin $θ$ = $\frac{1}{10}$ (as    $θ$   is very small),

when car is moving uphill :

JEE Main 2016 (Online) 9th April Morning Slot Physics - Work Power & Energy Question 78 English Explanation 1

P = f $\times$ u

=  (wsin $θ$ + f) $\times$ u

=   $(\frac{w}{10} + \frac{w}{20}) \times 10$

P = $\frac{3 w}{20} \times 10$ = $\frac{3 w}{2}$

When car is moving down hill :

JEE Main 2016 (Online) 9th April Morning Slot Physics - Work Power & Energy Question 78 English Explanation 2

$∴$     $\frac{P}{2}$ = (wsin $θ$ $-$ f) $\times$ v

$\Rightarrow$     $\frac{3 w}{4}$ = $(\frac{w}{10} - \frac{w}{20})$ $\times$ v

$\Rightarrow$     $\frac{w}{20} \times$ v = $\frac{3 w}{4}$

$\Rightarrow$    v = 15 m/s

17. (AIEEE 2005 )

A body of mass $m$ is accelerated uniformly from rest to a speed $v$ in a time $T .$ The instantaneous power delivered to the body as a function of time is given by

A. $\frac{m v^{2}}{T^{2}} . t^{2}$

B. $\frac{m v^{2}}{T^{2}} . t$

C. $\frac{1}{2} \frac{m v^{2}}{T^{2}} . t^{2}$

D. $\frac{1}{2} \frac{m v^{2}}{T^{2}} . t$

Correct Option is (B)

$\frac{1}{2} \frac{m v^{2}}{T^{2}} . t$

18. (AIEEE 2004)

A body of mass $^{'} m^{'},$ acceleration uniformly from rest to $^{'} {v_{1}}^{'}$ in time $T$ . The instantaneous power delivered to the body as a function of time is given by

A. $\frac{m v_{1} t^{2}}{T}$

B. $\frac{m v_{1}^{2} t}{T^{2}}$

C. $\frac{m v_{1} t}{T}$

D. $\frac{m v_{1}^{2} t}{T}$

Correct Option is (B)

Assume acceleration of body be $a$

$∴$ $v_{1} = 0 + a T \Rightarrow a = \frac{v_{1}}{T}$

$∴$ $v = a t \Rightarrow v = \frac{v_{1} t}{T}$

$P_{i n s t} = \vec{F} . \vec{v} = (m \vec{a}) . \vec{v}$

$= (\frac{m v_{1}}{T}) (\frac{v_{1} t}{T})$

$= m {(\frac{v_{1}}{T})}^{2} t$

19. (AIEEE 2003 )

A body is moved along a straight line by a machine delivering a constant power. The distance moved by the body in time $^{'} t^{'}$ is proportional to

A. $t^{3 / 4}$

B. $t^{3 / 2}$

C. $t^{1 / 4}$

D. $t^{1 / 2}$

Correct Option is (B)

We know that $F \times v =$ Power

According to the question, power is constant.

$∴$ $F \times v = c$ where $c =$ constant

$∴$ $m \frac{d v}{d t} \times v = c$ $($ $∴$ $F = m a = \frac{m d v}{d t})$

$∴$ $m \int_{0}^{v} v d v = c \int_{0}^{t} d t$ $∴$ $\frac{1}{2} m v^{2} = c t$

$∴$ $v = \sqrt{\frac{2 c}{m}} \times t^{1 / 2}$

$\frac{d x}{d t} = \sqrt{\frac{2 c}{m}} \times t^{1 / 2}$ where $v = \frac{d x}{d t}$

$∴$ $\int_{0}^{x} d x = \sqrt{\frac{2 c}{m}} \times \int_{0}^{t} t^{1 / 2} d t$
$x = \sqrt{\frac{2 c}{m}} \times \frac{2 t^{3 / 2}}{3} \Rightarrow x \propto t^{3 / 2}$