Question

A force F = 20 + 10y action 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. 25J
• B. 20J
• C. 30J
• D. 5J

Answer 1

The Correct Option is A

To calculate the work done by the force moving a particle from y = 0 to y = 1 \, \text{m}, we will use the concept of work done by a variable force. According to the physics of forces, when the force varies with position, the work done is given by the integral of the force over the distance.

The force acting on the particle in the y-direction is given by:

F = 20 + 10y

Here, the work done (W) by the force when a particle moves from y = 0 to y = 1 is calculated using the integral:

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

Let's evaluate the integral

• \int (20 + 10y) \, dy = \int 20 \, dy + \int 10y \, dy
• The integral of a constant is: \int 20 \, dy = 20y
• The integral of 10y is: \int 10y \, dy = 5y^2 (using the power rule for integration)

Now, we apply the limits y = 0 to y = 1:

• W = \left[20y + 5y^2\right]_{y=0}^{y=1}
• Calculating at y = 1: (20 \times 1 + 5 \times 1^2) = 20 + 5 = 25
• Calculating at y = 0: (20 \times 0 + 5 \times 0^2) = 0
• The work done is: 25 - 0 = 25 \, \text{Joules}

Therefore, the correct answer is 25 J.