Question

Power applied to a particle varies with time as P = (3t² - 2t+1) watt, where t is in second. Find the change in its kinetic energy between time t = 2 s and t = 4 s.

Answer 1

46

Answer 2

The change in kinetic energy is given by the integral of power with respect to time.

Given: Power, $P = (3t^2 - 2t + 1)$ watt Initial time, $t_1 = 2$ s Final time, $t_2 = 4$ s

The change in kinetic energy ($\Delta KE$) is:

$$\Delta KE = \int_{t_1}^{t_2} P \, dt$$ $$\Delta KE = \int_{2}^{4} (3t^2 - 2t + 1) \, dt$$

Integrate the expression:

$$\Delta KE = \left[ \frac{3t^3}{3} - \frac{2t^2}{2} + t \right]_{2}^{4}$$ $$\Delta KE = \left[ t^3 - t^2 + t \right]_{2}^{4}$$

Now, evaluate the definite integral by substituting the limits:

$$\Delta KE = [(4)^3 - (4)^2 + (4)] - [(2)^3 - (2)^2 + (2)]$$ $$\Delta KE = [64 - 16 + 4] - [8 - 4 + 2]$$ $$\Delta KE = [52] - [6]$$ $$\Delta KE = 46 \, \text{J}$$

The change in kinetic energy between t = 2 s and t = 4 s is 46 J.

Explanation of the solution: The change in kinetic energy is obtained by integrating the power function with respect to time over the given interval.

$$\Delta KE = \int_{t_1}^{t_2} P(t) \, dt = \int_{2}^{4} (3t^2 - 2t + 1) \, dt$$ $$\Delta KE = \left[ t^3 - t^2 + t \right]_{2}^{4} = (4^3 - 4^2 + 4) - (2^3 - 2^2 + 2) = (64 - 16 + 4) - (8 - 4 + 2) = 52 - 6 = 46 \, \text{J}$$