Question

A force \(F =\left(5+3 y^2\right)\) acts on a particle in the \(y\)-direction, where \(F\) is in newton and \(y\) is in meter The work done by the force during a displacement from \(y=2 m\) to \(y=5 m\) is___ \(J\).

Hint:

Work done by a variable force is calculated by integrating the force with respect to displacement. Ensure the force and displacement are in the same direction.

Answer 1

Correct Answer: 132

Step 1: Recall the Formula for Work Done

The work done (\(W\)) by a variable force \(F(y)\) over a displacement from \(y_1\) to \(y_2\) is given by the integral:

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

Step 2: Substitute the Given Force and Limits

In this case, \(F(y) = 5 + 3y^2\), \(y_1 = 2 \, \text{m}\), and \(y_2 = 5 \, \text{m}\). So,

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

Step 3: Evaluate the Integral

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

Conclusion: The work done by the force is \(132 \, \text{J}\).