MP_JEE_MAIN_09-01-19_S2 - Competishun – Live JEE & NEET Courses and Exam Prep

JEE MAIN 2019

09-01-19 S2

Question

A force acts on a 2 kg object so that its position is given as a function of time as $x=3 t^2+5$. What is the work done by this force in first 5 seconds?

Select the correct option:

950 J

900 J

850 J

875 J

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

$\begin{aligned} & V=\frac{d x}{d t}=6 \mathrm{t} \\ & V(\mathrm{t}=0)=0 \\ & V(\mathrm{t}=5 \mathrm{~s})=30 \mathrm{~m} / \mathrm{s} \\ & \Delta \mathrm{KE}=\frac{1}{2} 2 \times 30^2=900 \mathrm{~J}\end{aligned}$

Question Tags

JEE Main

Physics

Hard

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