Question

A body of mass 2 kg is accelerated from rest to a velocity. 20 ms$^{–1}$ in 5 s. What is the work done and power consumed.​

Answer 1

Answer:

Work done is 400 J and power consumed is 80 W.

Explanation:

Given:

• Mass (M) = 2 kg

• Initial velocity (u) = 0

• Final velocity (v) = 20 ms$^{–1}$

We know that,

Work done = 1/2 × m[v$^{2}$ – u$^{2}$ ]

Substitute the known values, we get

Work done = 1/2 × 2 × [20$^{2}$ – 0]

= 1/2 × 2 × 400 – 0

= 1/2 × 800

= 400 J

$\therefore$ Work done = 400 J

Now we will find power consumed

We know that,

Power = work/time

→ work = power × time

Now again substitute the known values, we get

power = work/time

= 400J/5s = 80 W

Thus, work done is 400 J and power consumed is 80 W.

Answer 2

Question :-

A body of mass 2 kg is accelerated from rest to a velocity. 20 ms$^{–1}$ in 5 s. What is the work done and power consumed.

Answer :-

• Power = 80 watt.
• Work = 400 Joule .

Given :-

(M) = 2 kg

(u) = 0

(v) = 20 ms$^{–1}$

Here ,

u = Initial velocity

v = Final velocity

m = mass of the body

Solution :-

Using formula of work done .

Work done = ½ × mv² – u²

Now ,

= ½ × 2 × 20² – 0

= ½ × 2 × 400 – 0

= ½ × 800

= 400 Joules

Hence we got

Work done = 400 Joules

So , now finding power

We know the formula of power :-

Power = work done / time

= 400J/5s

= 80 Watts

Therefore ,

• Work = 400 J
• Power = 80 W.

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