Question

A body of mass 2 kg is moving along x-direction such that its displacement as function of time is given by $x(t) = \alpha t^2 + \beta t + \gamma$ m, where $\alpha = 1$ m/s$^2$, $\beta = 1$ m/s and $\gamma = 1$ m. The work done on the body during the time interval $t = 2$ s to $t = 3$ s, is _________ J.

• A. 42
• B. 24
• C. 12
• D. 49

Hint:

Using the Work-Energy Theorem is often much faster than calculating force and integrating $F \cdot dx$, especially when displacement is a simple polynomial of time.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
According to the Work-Energy Theorem, the work done on a body by the net force is equal to the change in its kinetic energy. We first find the velocity of the body at the given time instants.

Step 2: Key Formula or Approach:
Velocity $v(t) = \frac{dx(t)}{dt}$.
Work done $W = \Delta K = \frac{1}{2} m (v_f^2 - v_i^2)$.

Step 3: Detailed Explanation:
Given $x(t) = t^2 + t + 1$.
The instantaneous velocity is:
\[ v(t) = \frac{d}{dt}(t^2 + t + 1) = 2t + 1 \]
Calculate velocities at $t = 2$ s and $t = 3$ s:
At $t_i = 2$ s: $v_i = 2(2) + 1 = 5$ m/s.
At $t_f = 3$ s: $v_f = 2(3) + 1 = 7$ m/s.
Calculate work done using $m = 2$ kg:
\[ W = \frac{1}{2} \cdot 2 \cdot (7^2 - 5^2) \]
\[ W = 49 - 25 = 24 \text{ J} \]

Step 4: Final Answer:
The work done is 24 J.