Question

A force\( F=20+10y \)acts on a particle in \(y-\)direction where \(F\) is in newton and y in meter. Work done by this force to move the particle from \(y=0\) to \( y=1\) m is: 

• A. \(30J\)
• B. \(5J\)
• C. \(25J\)
• D. \(20J\)

Answer 1

The Correct Option is C

To find the work done by the force \( F(y) = 20 + 10y \) while moving the particle from \( y = 0 \) to \( y = 1 \) meter, we need to use the concept of work done by a variable force. The work done by a force over a distance is given by the integral of the force with respect to the displacement:

The formula for work done \( W \) is:

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

Here, the limits of integration are from \( y_1 = 0 \) to \( y_2 = 1 \).

Substitute the expression for \( F(y) \):

W = \int_{0}^{1} (20 + 10y) \, dy

Now, let's evaluate the integral:

W = \int_{0}^{1} 20 \, dy + \int_{0}^{1} 10y \, dy

Evaluate each integral separately:

• \int_{0}^{1} 20 \, dy = 20y \big|_{0}^{1} = 20(1) - 20(0) = 20
• \int_{0}^{1} 10y \, dy = 10\left(\frac{y^2}{2}\right) \big|_{0}^{1} = 10\left(\frac{1^2}{2}\right) - 10\left(\frac{0^2}{2}\right) = 5

Sum these to find the total work done:

W = 20 + 5 = 25 \text{ J}

Thus, the work done by the force to move the particle from \( y = 0 \) to \( y = 1 \) m is 25 J.