PYQ NEET- Work Energy And Power L-5

Question:#

A force $\vec{F}=(2+3 x) \hat{i}$ acts on a particle in the $X$ direction where $\mathrm{F}$ is in newton and $\mathrm{X}$ is in meter. The work done by this force during a displacement from $X=0 \quad X=4 \mathrm{~m}$, is J.

Answer:#

To find the work done by a force during a displacement, we can use the formula: $$ W=\int_{x_1}^{x_2} \vec{F} \cdot d \vec{x} $$

Here, the force is given by $\vec{F}=(2+3 x) \hat{i}$, and we need to find the work done during a displacement from $x=0$ to $x=4 \mathrm{~m}$. Since the force is only in the $x$ direction, we can write the integral as: $$ W=\int_0^4(2+3 x) d x $$

Now we can integrate the function with respect to $x$ : $$ \begin{aligned} & W=\int_0^4(2+3 x) d x=\int_0^4 2 d x+\int_0^4 3 x d x \ & W=[2 x]_0^4+\left[\frac{3}{2} x^2\right]_0^4 \end{aligned} $$

Now we can plug in the limits of integration: $$ \begin{aligned} & W=(2 \cdot 4-2 \cdot 0)+\left(\frac{3}{2} \cdot 4^2-\frac{3}{2} \cdot 0^2\right) \ & W=8+24 \ & W=32 \mathrm{~J} \end{aligned} $$

So the work done by the force during the displacement from $x=0$ to $x=4 \mathrm{~m}$ is 32 Joules.