Question

Compute the power required by a men to move a 10 kg mass to change the speed from 10 m/s to 20m/s in 5 second

Answer 1

Answer:

Power delivered will be 300 watt.

Answer 2

Answer:

• Power of the Man (P) = 300 Watts.

Given:

1. Mass of the Man (M) = 10 Kg.
2. Initial velocity (u) = 10 m/s.
3. Final velocity (v) = 20 m/s.
4. Time Period (t) = 5 Seconds.

Explanation:

$\rule{300}{1.5}$

Kinetic energy:-

Kinetic energy is the energy which is Possessed by the virtue of its motion.

Formula:-

K.E = ½Mv²

$\bold{Here}\begin{cases}{\sf{M = Mass \: of \: the\: Body}} \\ \sf{v = Velocity\: of\: the\: body}\end{cases}$

Power:-

Power is Defined as rate of doing work (or) Energy.

Formula:-

Power = E/T.

$\bold{Here}\begin{cases}{\sf{T = Time \: Period }} \\ \sf{E = Energy \: of\: the\: body} \\ \sf{E = Potential \: Energy \: (or) \: Kinetic\: Energy}\end{cases}$

$\rule{300}{1.5}$

$\rule{300}{1.5}$

Kinetic energy (E) :-

$\large{\boxed{\tt E = \dfrac{1}{2}m(v^2 - u^2)}}$

Substituting the values,

$\large{\tt \hookrightarrow E = \dfrac{1}{2} \times 10 \times [(20)^2 - (10)^2]}$

$\large{\tt \hookrightarrow E = \dfrac{1}{\cancel{2}} \times \cancel{10} \times [400 - 100]}$

$\large{\tt \hookrightarrow E = 5 \times [400 - 100]}$

$\large{\tt \hookrightarrow E = 5 \times 300}$

$\large{\boxed{\tt E = 1500 \: J}}$

So, the Kinetic energy of the body is 1500 Joules.

$\rule{300}{1.5}$

$\rule{300}{1.5}$

Power (P) :-

$\large{\boxed{\tt P = \dfrac{E}{t}}}$

Substituting the values,

$\large{\tt \hookrightarrow P = \dfrac{1500}{5}}$

$\large{\tt \hookrightarrow P = \dfrac{5 \times 300}{5}}$

$\large{\tt \hookrightarrow P = \dfrac{\cancel{5} \times 300}{\cancel{5}}}$

$\large{\tt \hookrightarrow P = 1 \times 300}$

$\huge{\boxed{\boxed{\tt P = 300 \: W}}}$

So, the Power performed by the Man is 300 Watts.

$\rule{300}{1.5}$