Question

what work a boy of mass 50kg will do in order to increase running speed from 9km/h to 18km/h​

Answer 1

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

m = 50 Kg.

v = 18 Km/h

$\large{ 18 Km/h = \frac{5}{18} \times18 \: m/s}$

v = 5 m/s.

u = 9 Km/h

$\large{ 9 Km/h = \frac{5}{18} \times9 \: m/s}$

u = 2.5 m/s.

$\huge{\bold{\underline{Explanation:-}}}$

As the mass and velocity are given

We need to use the formula.

$\large{\bold{W = \frac{1}{2}m(v^2 - u^2)}}$

Substituting the values.

$\large{ \implies W = \frac{1}{2} \times 50 (5^2 - 2.5^2)}$

$\large{ \implies W = \frac{1}{ \cancel{2}} \times \cancel{50} \times (25 - 6.25)}$

$\large{ \implies W = 25 \times 18.75 }$

$\huge{\boxed{\boxed{W = 468.75 \: J}}}$

Additional formulas:-

$\large{\bold{ \star W = mgh}}$

$\large{\bold{ \star W = F.s}}$

$\large{\bold{ \star W = \frac{1}{2}Kx^2}}$

Answer 2

Answer:-

Work done = 468.75J

Explanation:-

Given:-

• m = 50kg
• $v_1 = 9kmph = 2.5m{s}^{-1}$
• $v_2 = 18kmph = 5m{s}^{-1}$

To Find:-

Work done by the boy

Solution:-

Work done = ΔK.E. of the boy

$ΔK.E. = K.E_2 - K.E _1$

$ΔK.E. = \dfrac{1}{2}m{v_2}^{2} - \dfrac{1}{2}m{v_1}^{2}$

$ΔK.E. = \dfrac{1}{2}m({v_2}^{2} -{v_1}^{2})$

$ΔK.E. = \dfrac{1}{2} \times 50({5}^{2} -{2.5}^{2})$

$ΔK.E. = 25 (25 -6.25)$

$ΔK.E. = 25 (18.75)$

$ΔK.E. = 468.75J$

$\bold{W = 468.75J}$