Question 23

A force $$F = (5 + 3y^{2})$$ 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.

Name: A force F = (5 + 3y^{2}) 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.
Uploaded: 2026-05-13T11:46:52.607096+05:30
Duration: 59 s
Description: A force F = (5 + 3y^{2}) 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.

Correct Answer: 132

Solution

$$W = \int_{y_i}^{y_f} F(y) \, dy$$

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

$$W = \left[ 5y + \frac{3y^3}{3} \right]_2^5$$

$$W = \left[ 5y + y^3 \right]_2^5$$

$$W = 132 \text{ J}$$

Get AI Help

Video Solution

Create a FREE account and get:

Free JEE Mains Previous Papers PDF
Take JEE Mains paper tests

A force $$F = (5 + 3y^{2})$$ 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.

Solution

Get AI Help

Video Solution

Download resources