PYQ NEET- Work Energy And Power L-1

Question:#

A block of mass $10 \mathrm{~kg}$ is moving along $\mathrm{x}$-axis under the action of force $\mathrm{F}=5 \mathrm{xN}$. The work done by the force in moving the block from $x=2 m$ to $4 \mathrm{~m}$ will be J.

Answer:#

To calculate the work done by the force $F=5 x$ in moving the block from $x=2 m$ to $x=4 m$, we can use the formula for work done by a variable force: $$ W=\int_{x_1}^{x_2} F(x) d x $$

In this case, $F(x)=5 x, x_1=2 m$, and $x_2=4 m$. Now, we can substitute these values into the formula and evaluate the integral: $$ W=\int_2^4 5 x d x $$

To evaluate the integral, we find the antiderivative of $5 x$ : $$ \int 5 x d x=\frac{5}{2} x^2+C $$

Now, we can find the work done by evaluating the antiderivative at the limits of integration: $$ \begin{aligned} & W=\left[\frac{5}{2} x^2\right]_2^4=\frac{5}{2}\left(4^2\right)-\frac{5}{2}\left(2^2\right) \ & W=\frac{5}{2}(16)-\frac{5}{2}(4)=40-10=30 J \end{aligned} $$

The work done by the force in moving the block from $x=2 m$ to $x=4 m$ is $30 \mathrm{~J}$.