- Displacement, \(d = 2\text{ m}\)

- Force, \(F = 20\text{ N}\)

To find:

- Work done, \(W\)

Solution:

The work done on the TV set is given by:

$$W = Fd\cos\theta$$

Where \(\theta\) is the angle between the force and displacement vectors. In this case, since the force and displacement are in the same direction, \(\theta = 0\degree\). Therefore, \(\cos\theta = 1\).

Substituting the given values, we get:

$$W = (20\text{ N})(2\text{ m})(1) = \boxed{40\text{ J}}$$

Therefore, the work done on the TV set is 40 J.