JEE Physics 2023 Question with Solution

Given: The force is $F(y) = 5 + 3y^2$ and the displacement is from $y = 2 \, \text{m}$ to $y = 5 \, \text{m}$.

Find: The work done $W$ by this variable force.

For a variable force acting along the direction of displacement,

$$W = \int_{y_1}^{y_2} F(y) \, dy$$

Substituting the given force and limits,

$$W = \int_{2}^{5} (5 + 3y^2) \, dy$$

Now evaluate the integral,

$$W = \left[5y + \frac{3y^3}{3}\right]_2^5 = \left[5y + y^3\right]_2^5$$ $$W = (5(5) + 5^3) - (5(2) + 2^3)$$ $$W = (25 + 125) - (10 + 8)$$ $$W = 150 - 18 = 132 \, \text{J}$$

Therefore, the work done is $132 \, \text{J}$.

Given: $F(y) = 5 + 3y^2$.

Find: Work done from $y=2$ to $y=5$.

The required expression is

$$W = \int_{2}^{5} (5 + 3y^2) \, dy$$

Integrate term by term:

$$\int 5 \, dy = 5y$$ $$\int 3y^2 \, dy = y^3$$

So,

$$W = \left[5y + y^3\right]_2^5$$

At $y=5$,

$$5(5) + 5^3 = 25 + 125 = 150$$

At $y=2$,

$$5(2) + 2^3 = 10 + 8 = 18$$

Hence,

$$W = 150 - 18 = 132$$

Therefore, the numerical value of the answer is 132.