JEE Main Work, Energy and Power PYQ

21. (AIEEE 2004 )

A force $\vec{F} = (5 \vec{i} + 3 \vec{j} + 2 \vec{k}) N$ is applied over a particle which displaces it from its origin to the point $\vec{r} = (2 \vec{i} - \vec{j}) m .$ The work done on the particle in joules is

A. $+ 10$

B. $+ 7$

C. $- 7$

D. $+ 13$

Correct Option is (B)

The work done by a force on a particle is given by the dot product of the force and the displacement vector of the particle: $W = \vec{F} \cdot \vec{r}$

We can substitute the given vectors into this expression:

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

$= (5 \hat{i} + 3 \hat{j} + 2 \hat{k}) . (2 \hat{i} - \hat{j})$

$= 10 - 3 = 7$ J

22. (AIEEE 2003 )

A spring of spring constant $5 \times 10^{3} N / m$ is stretched initially by $5$ $c m$ from the unstretched position. Then the work required to stretch it further by another $5$ $c m$ is

A. $12.50$ $N$ - $m$

B. $18.75$ $N$ - $m$

C. $25.00$ $N$ - $m$

D. $625$ $N$ - $m$

Correct Option is (B)

Given $k = 5 \times 10^{3} N / m$

Work done when a spring stretched from x₁ cm to x₂ cm,

$W = \frac{1}{2} k (x_{2}^{2} - x_{1}^{2})$

$= \frac{1}{2} \times 5 \times 10^{3} [{(0.1)}^{2} - {(0.05)}^{2}]$

$= \frac{5000}{2} \times 0.15 \times 0.05 = 18.75 N m$

23. (AIEEE 2002 )

A spring of force constant $800$ $N / m$ has an extension of $5$ $c m .$ The work done in extending it from $5$ $c m$ to $15$ $c m$ is

A. $16 J$

B. $8 J$

C. $32 J$

D. $24 J$

Correct Option is (B)

When we extend the spring by $d x$ then the work done

$d W = k x d x$

Applying integration both sides we get,

$∴$ $W = k \int_{0.05}^{0.15} x d x$

$= \frac{800}{2} [{(0.15)}^{2} - {(0.05)}^{2}]$

$= 8 J$