Question

a force of 5 n acts on a 15 kg body initially at rest the work done by the force during the first second of motion of a body is​

Answer 1

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Gicen:-

Force = 5N

Mass = 15kg

To find:-

Work done by force during the 1st sec. of motion.

Solution :-

a = $\bf\large\frac{F}{m}$

a = $\bf\large\frac{5}{15}$

a = $\bf\large\frac{1}{3}$

We know,

Work = Force × displacement

W = Fs

Work = 5 × $\bf\large\frac{1}{2}$ at²

$\bf\underbrace{s=\:ut +1/2 at^2}$

Work = 5 × $\bf\large\frac{1}{2}$ × $\bf\large\frac{1}{3}$× 1²

Work = 2.5 × $\bf\large\frac{1}{3}$ × 1²

$\bf{\large{\boxed{\underline{\rm{\pink{<strong><u> </u></strong><strong><u>Work </u></strong><strong><u>=</u></strong><strong><u> </u></strong><strong><u>0.</u></strong><strong><u>8</u></strong><strong><u>3</u></strong><strong><u>3</u></strong><strong><u>3</u></strong><strong><u> </u></strong><strong><u>J</u></strong>}}}}}}$

Note:- The fractional form of the answer is 5/6 J. However, on dividing further the answer would be 0.8333J. No worry about the answer mine is too correct! :)