A force $F = (20 + 10y)$ acts on a particle in the $y$-direction where $F$ is in Newton and $y$ is in metre. T

Question

A force $F = (20 + 10y)$ acts on a particle in the $y$-direction where $F$ is in Newton and $y$ is in metre. The work done by this force to move the particle from $y = 0$ to $y = 1$ m is:

Accepted Answer

When a force varies with position, the work done ($W$) is calculated by integrating the force over the displacement . For a force $F(y)$ acting in the $y$-direction, the formula is:

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

Given the force $F = (20 + 10y)$ and the limits from $y = 0$ to $y = 1$ m:

1. **Set up the integral**: 
   $$W = \int_{0}^{1} (20 + 10y) dy$$

2. **Integrate the expression**: 
   $$W = \left[ 20y + \frac{10y^2}{2} ight]_{0}^{1} = [20y + 5y^2]_{0}^{1}$$

3. **Apply the limits**: 
   $$W = (20(1) + 5(1)^2) - (20(0) + 5(0)^2)$$
   $$W = 20 + 5 = 25 	ext{ J}$$

Thus, the total work done is **25 J**.

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