JEE Main Work, Energy and Power PYQ

11.(JEE Main 2021 (Online) 27th August Morning Shift )

Two persons A and B perform same amount of work in moving a body through a certain distance d with application of forces acting at angle 45 $^{\circ}$ and 60 $^{\circ}$ with the direction of displacement respectively. The ratio of force applied by person A to the force applied by person B is $\frac{1}{\sqrt{x}}$ . The value of x is .................... .

Correct answer is (2)

Given W_A = W_B

F_Ad cos45 $^{\circ}$ = F_Bd cos60 $^{\circ}$

$F_{A} \times \frac{1}{\sqrt{2}} = F_{B} \times \frac{1}{2}$

$\frac{F_{A}}{F_{B}} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$

x = 2

12.(JEE Main 2021 (Online) 25th July Evening Shift )

A force of F = (5y + 20) $\hat{j}$ N acts on a particle. The work done by this force when the particle is moved from y = 0 m to y = 10 m is ___________ J.

Correct answer is (450)

F = (5y + 20) $\hat{j}$

$W = \int F d y = \int_{0}^{10} (5 y + 20) d y$

$= {(\frac{5 y^{2}}{2} + 20 y)}_{0}^{10}$

$= \frac{5}{2} \times 100 + 20 \times 10$

$= 250 + 200 = 450$ J

13. (JEE Main 2020 (Online) 4th September Evening Slot )

A person pushes a box on a rough horizontal plateform surface. He applies a force of 200 N over a distance of 15 m. Thereafter, he gets progressively tired and his applied force reduces linearly with distance to 100 N. The total distance through which the box has been moved is 30 m. What is the work done by the person during the total movement of the box?

A. 5690 J

B. 5250 J

C. 2780 J

D. 3280 J

Correct Option is (B)

JEE Main 2020 (Online) 4th September Evening Slot Physics - Work Power & Energy Question 61 English Explanation

Work done = area of ABCEO

= area of trapezium ABCD + area of rectangle ODCE

= $\frac{1}{2}$ $\times$ 45 $\times$ 30 + 100 $\times$ 30 = 5250J

14. (JEE Main 2020 (Online) 9th January Morning Slot )

Consider a force $\vec{F} = - x \hat{i} + y \hat{j}$ . The work done by this force in moving a particle from point A(1, 0) to B(0, 1) along the line segment is : (all quantities are in SI units) JEE Main 2020 (Online) 9th January Morning Slot Physics - Work Power & Energy Question 66 English

A. 2

B. $\frac{1}{2}$

C. 1

D. $\frac{3}{2}$

Correct Option is (C)

W = $\int \vec{F} . d \vec{r}$

= $\int (- x \hat{i} + y \hat{j}) . (d x \hat{i} + d y \hat{j})$

= $- \int_{1}^{0} x d x + \int_{0}^{1} y d y$

= $\frac{1}{2} + \frac{1}{2}$ = 1 J

15. (JEE Main 2020 (Online) 7th January Morning Slot )

A 60 HP electric motor lifts an elevator having a maximum total load capacity of 2000 kg. If the frictional force on the elevator is 4000 N, the speed of the elevator at full load is close to :
(1 HP = 746 W, g = 10 ms^-2)

A. 1.5 ms^-1

B. 1.7 ms^-1

C. 2.0 ms^-1

D. 1.9 ms^-1

Correct Option is (D)

F = mg + f

F = 20000 + 4000 = 24000 N

We know, Power(P) = Fv

$\Rightarrow$ v = $\frac{P}{F}$ = $\frac{60 \times 746}{24000}$

$\Rightarrow$ v $\approx$ 1.9 m/s