Question

A block of mass 1 kg is free to move along the x-axis. It is at rest and from time $t=0$ onwards it is subjected to a time-dependent force $F(t)$ in the x-direction. The force $F(t)$ varies with $t$ as shown in figure. The kinetic energy of the block at $t=4$ s is

• A. 2 J
• B. 3 J
• C. 0 J
• D. 4 J
• E.

Hint:

For any $F-t$ graph, if the total area above the time axis equals the total area below it, the net impulse is zero, meaning the final velocity will be equal to the initial velocity.

Answer 1

The Correct Option is D

Concept:
The area under a Force-time ($F-t$) graph represents the Impulse, which is equal to the change in momentum ($\Delta p$). \[ \text{Impulse} = \int F \, dt = \Delta p = m(v_f - v_i) \]

Step 1: Calculate the net area under the $F-t$ graph from 0 to 4 s.
Based on the graph, the shape consists of four triangles with equal bases (1 s) and equal magnitude heights (2 N): [label=\alph*), itemsep=8pt]
• Area 1 (0 to 1 s): $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 1 \times 2 = +1$ Ns.
• Area 2 (1 to 2 s): $\frac{1}{2} \times 1 \times (-2) = -1$ Ns.
• Area 3 (2 to 3 s): $\frac{1}{2} \times 1 \times (-2) = -1$ Ns.
• Area 4 (3 to 4 s): $\frac{1}{2} \times 1 \times 2 = +1$ Ns.

Step 2: Determine final velocity and Kinetic Energy.
Total Impulse = $1 - 1 - 1 + 1 = 0$ Ns.
Since Impulse = $m(v_f - v_i)$ and the block starts from rest ($v_i = 0$): \[ 0 = 1(v_f - 0) \implies v_f = 0 \text{ m/s} \] Kinetic Energy ($K.E.$) at $t = 4$ s is $\frac{1}{2} m v_f^2 = \frac{1}{2} (1)(0)^2 = 0$ J.