Question

A block of mass 10 kg is put gently on a belt-conveyor system of infinite length at t = 0 sec, which is moving with constant speed 20 m/sec rightward at all time, irrespectively of any situation by means of a motor-system as shown in the figure. A constant force of magnitude 15 N is applied on the block continuously during its motion.

• A. 1250 Joule
• B. 2500 Joule
• C. -1250 Joule
• D. Zero

Answer 1

Answer: (A)

1250 Joule

Answer 2

1. Identify forces and motion phases: The block (10 kg) starts from rest on a belt moving at 20 m/s. A 15 N force acts rightward. Kinetic friction acts initially rightward ($f_k = \mu_k N = 0.25 \times 10 \times 10 = 25$ N).
2. Calculate initial acceleration: Net force $F_{net} = 15 \text{ N} + 25 \text{ N} = 40 \text{ N}$. Acceleration $a = F_{net}/m = 40/10 = 4 \text{ m/s}^2$.
3. Time to reach belt speed: The block accelerates until its speed equals the belt's speed (20 m/s). $v = u + at \implies 20 = 0 + 4t_s \implies t_s = 5$ sec.
4. Work done by kinetic friction: Kinetic friction acts only during $0 \le t \le 5$ sec. Displacement of block $s_b = ut_s + \frac{1}{2}at_s^2 = 0 + \frac{1}{2}(4)(5^2) = 50$ m. Work done $W_k = f_k \times s_b = 25 \text{ N} \times 50 \text{ m} = 1250$ Joule.