Search any question & find its solution
Question:
Answered & Verified by Expert
At a place where the acceleration due to gravity is ${ }^{10 \mathrm{msec}^{-2}}$ a force of $5 \mathrm{~kg}-\mathrm{wt}$ acts on a body of mass $10 \mathrm{~kg}$ initially at rest. The velocity of the body after 4 second is
Options:
Solution:
2169 Upvotes
Verified Answer
The correct answer is:
$20 \mathrm{msec}^{-1}$
$\text { Acceleration }=\frac{\text { Force }}{\text { Mass }}=\frac{50 \mathrm{~N}}{10 \mathrm{~kg}}=5 \mathrm{~m} / \mathrm{s}^2$
From $v=u+a t=0+5 \times 4=20 \mathrm{~m} / \mathrm{s}$
From $v=u+a t=0+5 \times 4=20 \mathrm{~m} / \mathrm{s}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.