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

To find the work done by the force \(F = 5 + 3y^2\) as it moves a particle from \(y=2\ \text m\) to \(y=5\ \text m\), we need to calculate the definite integral of the force function with respect to \(y\) over the interval from 2 to 5.

The work done, \(W\), by a variable force is given by:

\(W=\int_{y_1}^{y_2} F\ \text dy = \int_2^5 (5 + 3y^2)\ \text dy\)

Calculating the integral:

\(\int (5 + 3y^2)\ \text dy = \int 5\ \text dy + \int 3y^2\ \text dy\)

The integral of each term is:

\(\int 5\ \text dy = 5y\) and \(\int 3y^2\ \text dy = y^3\)

Thus, the antiderivative of \((5 + 3y^2)\) is:

\(5y + y^3\)

Evaluating this from \(y=2\) to \(y=5\):

\([5(5) + (5)^3] - [5(2) + (2)^3]\)

Calculating each part:

• \(5(5)=25\) and (5)^3=125\)
• \(5(2)=10\) and (2)^3=8\)

Substituting these back:

\((25 + 125) - (10 + 8) = 150 - 18 = 132\)

Thus, the work done by the force is \(132\ \text J\).

Checking against the given range (132, 132), the computed work of \(132\ \text J\) is indeed within the expected range.