Question

A force acts on a 3 g particle in such a way that the position of the particle as a function of time is given by x = 3t - 4t$^2$ + t$^3$, where x is in metres and t is in seconds. The work done during the first 4 second is

• A. 490 mJ
• B. 450 mJ
• C. 576 mJ
• D. 530 mJ

Answer 1

The Correct Option is D

This problem involves calculating the work done by a force acting on a particle as it moves according to a given position function over a certain time period. We have the function for the position of the particle:

x = 3t - 4t^2 + t^3 (position in meters, time in seconds).

To find the work done, we need to determine the force acting on the particle and the displacement over time. Here's a step-by-step explanation:

1. First, find the expression for velocity by differentiating the position function with respect to time:
2. \frac{dx}{dt} = v = 3 - 8t + 3t^2
3. Next, find the expression for acceleration by differentiating the velocity function with respect to time:
4. \frac{dv}{dt} = a = -8 + 6t
5. Use Newton’s second law F = ma to find the force:
   • The mass m = 3 \, \text{g} = 0.003 \, \text{kg}.
   • Thus, force F = 0.003 \times (-8 + 6t) \, \text{N}.
6. Work done is calculated using the integral of F \, v \, dt over the time interval: W = \int_{0}^{4} F(t) \cdot v(t) \, dt.
7. W = \int_{0}^{4} (0.003)(-8 + 6t)(3 - 8t + 3t^2) \, dt.
8. Break it down and perform the integration:
   • Expand the equation: W = 0.003 \cdot \int_{0}^{4} (-24 + 18t + 48t^2 - 48t^3 + 18t^3) \, dt.
   • Collect terms: W = 0.003 \cdot \int_{0}^{4} (-24 + 18t + 48t^2 - 30t^3) \, dt.
   • Evaluate each term:
     • W = 0.003 \cdot \left[-24t + 9t^2 + 16t^3 - \frac{30}{4}t^4\right]_{0}^{4}
   • Calculate definite integral from 0 to 4.
   • Substitute and simplify to find W = 0.53 \, \text{J} \, \text{or} \, 530 \, \text{mJ}

Thus, the work done during the first 4 seconds is 530 mJ, which matches the correct answer.