A 70 kg box is initially at rest at the bottom of a ROUGH plane inclined at an angle of 30° to the horizontal - NSC Physical Sciences - Question 5 - 2019 - Paper 1

The mechanical energy of the system is not conserved due to the presence of friction, which dissipates energy as heat.

The free-body diagram should include:

• Weight (W) acting downwards: W = mg = 70 kg * 9.8 m/s²
• Normal force (N) perpendicular to the plane.
• Tension (T) acting parallel to the plane along the rope upward.
• Frictional force (f) acting down the plane against the direction of motion.

The work done by the frictional force is calculated using the formula: $W = f imes d imes ext{cos}( heta)$ where:

• f = 178.22 N (frictional force)
• d = 4 m (distance)
• heta = 180° (the frictional force is opposite to the direction of motion). Thus, $W = 178.22 imes 4 imes ext{cos}(180°) = -712.88 ext{ J}$

Applying the work-energy principle: The net work done on the box equals the change in kinetic energy. The equation is: $W_{net} = riangle KE = KE_{final} - KE_{initial}$ Assuming the box starts from rest, the initial kinetic energy is 0. The net work done is given by: $W_{net} = T - f = 700 N - 178.22 N = 521.78 N$ Then, $KE_{final} = W_{net} imes d = 521.78 N imes 4 m = 2087.12 J$ Using the kinetic energy formula, where ( KE = \frac{1}{2} mv^2 ): $2087.12 = \frac{1}{2} \times 70 v^2$ Solving for v gives: $v = \sqrt{\frac{2087.12 \times 2}{70}} \approx 5.42 ext{ m/s}$

When the box moves up the incline, friction does negative work, which we calculated as -712.88 J. When the box slides down, friction will do positive work, equal to the opposite of the frictional force, also -178.22 N over 4 m. Therefore: Total work done by friction during the entire process: $W_{total} = -712.88 J + 712.88 J = 0 J$.