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EEJ MAIN Physics QUESTION #7028
Question 1
A force acts on an object of mass $2\text{ kg}$ such that its position as a function of time is $x = 3t^2 + 5$. Calculate the work performed by this force during the first $5$ seconds of motion.
  • $850\text{ J}$
  • $950\text{ J}$
  • $875\text{ J}$
  • $900\text{ J}$✔️
Correct Answer Explanation
Work done is equal to the change in kinetic energy ($\Delta KE$). First, find velocity $v = \frac{dx}{dt} = 6t$. At $t=0$, $v_i = 0$. At $t=5$, $v_f = 6(5) = 30\text{ m/s}$. $\Delta KE = \frac{1}{2}m(v_f^2 - v_i^2) = \frac{1}{2}(2)(30^2 - 0) = 900\text{ J}$.