Work, Energy and Power: Numerical Problems Set (1)

Horizontal component: $F_x = 20\cos 30^\circ = 10\sqrt{3}$ N. Vertical component: $F_y = 20\sin 30^\circ = 10$ N (upward).

Vertical equilibrium: $N + F_y = mg \Rightarrow N = 20 - 10 = 10$ N. The applied force lifts the block partially, so $N$ is less than $mg$.

Friction opposes motion, so it acts backwards.

• Applied force: $W_F = F\cos 30^\circ \times d = 10\sqrt{3} \times 5 = 50\sqrt{3} \approx 86.6$ J
• Friction: $W_f = -f \times d = -2 \times 5 = -10$ J
• Gravity: $W_g = 0$ (perpendicular to motion)
• Normal: $W_N = 0$ (perpendicular to motion)

Final answer: $W_F \approx 86.6$ J, $W_f = -10$ J, $W_g = W_N = 0$, $W_{net} \approx 76.6$ J.

Why This Works

Work done by a force equals $F \cdot d \cos\theta$ where $\theta$ is the angle between force and displacement. Forces perpendicular to motion do zero work — that is why $N$ and $mg$ contribute nothing.

The trick most students miss is recomputing $N$ when the applied force has a vertical component. If you assume $N = mg$ here, your friction will be wrong by 50%.

Alternative Method

Use the work-energy theorem to verify. If you also compute final kinetic energy from $W_{net} = \Delta KE$, you can check the block’s final speed assuming it started at rest: $v = \sqrt{2 W_{net}/m} \approx 8.75$ m/s.