Numerical 1: Work Done Against Friction on Inclined Plane

Problem: A 2 kg block is pushed 5 m along a rough horizontal surface (μₖ = 0.3) by a force F = 20 N at 30° to the horizontal (g = 10 m/$s^{2}$). Find the net work done.

Solution:

Step 1 — Normal force (taking vertical equilibrium):

$N = mg - F\sin 30°$ $N = (2\ \text{kg})(10\ \text{m/s}^2) - (20\ \text{N})(0.5)$ $N = 20\ \text{N} - 10\ \text{N} = 10\ \text{N}$

Step 2 — Kinetic friction force:

$f_k = \mu_k N = (0.3)(10\ \text{N}) = 3\ \text{N}$

Step 3 — Work done by applied force:

$W_F = F\cos 30° \times d = (20\ \text{N})(0.866)(5\ \text{m}) = 86.6\ \text{N·m} = 86.6\ \text{J}$

Step 4 — Work done by friction (opposing motion, θ = 180°):

$W_f = -f_k \times d = -(3\ \text{N})(5\ \text{m}) = -15\ \text{J}$

Step 5 — Work done by gravity (motion is horizontal):

$W_g = mg \times d \times \cos 90° = 0\ \text{J}$

Step 6 — Work done by normal force (N perpendicular to motion):

$W_N = N \times d \times \cos 90° = 0\ \text{J}$

Step 7 — Net work:

$W_{\text{net}} = W_F + W_f + W_g + W_N = 86.6\ \text{J} - 15\ \text{J} + 0\ \text{J} + 0\ \text{J} = \boxed{71.6\ \text{J}}$

Step 8 — Verify using $\Delta K$E: By work-energy theorem, the block's KE increases by 71.6 J. If it started from rest: ½(2)$v^{2}$ = 71.6; v = 8.46 m/s.