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where $W$ is the net work done and $\Delta KE$ is the change in kinetic energy. Let's consider some problems and their solutions:
A force of 10 N is applied to an object, causing it to move 5 m in the direction of the force. Calculate the work done.
$$W = 10 \cdot 5 = 50 J$$
Now, we can calculate the work done:
$$W = \Delta KE$$
An object of mass 2 kg is moving with a velocity of 4 m/s. A force of 5 N is applied in the opposite direction of motion, causing it to slow down. Calculate the work done by the force.
where $v = 0$ (final velocity), $u = 4$ m/s (initial velocity), $a = -\frac{5}{2}$ m/s$^2$ (acceleration), and $s$ is the displacement. dc pandey physics class 11 pdf download part 1 work
$$s = \frac{u^2}{2a} = \frac{4^2}{2 \cdot \frac{5}{2}} = \frac{16}{5} m$$