Solve the following :

Question:

A block of mass $2.00 \mathrm{~kg}$ moving at a speed of $10.0 \mathrm{~m} / \mathrm{s}$ accelerates at $3.00 \mathrm{~m} / \mathrm{s}^{2}$ for $5.00 \mathrm{~s}$. Compute its final kinetic energy.

Solution:

Final speed $v=u+$ at

$=10+3 \times 5$

$v=25 \mathrm{~m} / \mathrm{s}$

Now,

$K . E .=\frac{1}{2} m v^{2}$

$=\frac{1}{2} \times 2 \times 25^{2}$

K.E. $=625 \mathrm{~J}$

Leave a comment

Close

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now