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

To find the work done on the body during the time interval from t = 2 s to t = 3 s, we need to determine the change in kinetic energy, since work done is equal to the change in kinetic energy.

First, we have the displacement function defined as:

x(t) = \alpha t^2 + \beta t + \gamma, where \alpha = 1 \, \text{m/s}^2, \beta = 1 \, \text{m/s}, and \gamma = 1 \, \text{m}.

The velocity is obtained by differentiating the displacement function with respect to time:

v(t) = \frac{dx}{dt} = 2\alpha t + \beta = 2 \times 1 \, t + 1 = 2t + 1 \, \text{m/s}.

Next, calculate the velocity at t = 2 s:

v(2) = 2 \times 2 + 1 = 5 \, \text{m/s}.

Now, calculate the velocity at t = 3 s:

v(3) = 2 \times 3 + 1 = 7 \, \text{m/s}.

The kinetic energy at any time t is given by:

KE = \frac{1}{2} m v^2, where m = 2 \, \text{kg}.

Thus, the kinetic energy at t = 2 s is:

KE_2 = \frac{1}{2} \times 2 \times (5)^2 = 25 \, \text{J}.

The kinetic energy at t = 3 s is:

KE_3 = \frac{1}{2} \times 2 \times (7)^2 = 49 \, \text{J}.

Therefore, the work done on the body is the change in kinetic energy:

\Delta KE = KE_3 - KE_2 = 49 \, \text{J} - 25 \, \text{J} = 24 \, \text{J}.

This confirms that the work done on the body from t = 2 s to t = 3 s is 24 J, matching the correct answer.