Question

A particle of mass 0.2 kg is moving in one-dimension under a force that delivers a constant power 0.5
W to the particle. If the initial speed (in ms-1) of the particle is zero, then speed (in ms-l) after 5 s is ​

Answer 1

Given :-

• Mass = 0.2 kg
• Power = 0.5 W
• Time = 5 s
• u = 0

To find :-

• Speed of body after 5 seconds

Formula Implemented:-

P = $\sf\dfrac{W}{T}$

Here work done = K.E

P = $\sf\dfrac{mv²}{2T}$

Plugging values :-

0.5 = $\sf\dfrac{0.2 v²}{10}$

Do cross multiplication

0.5 (10) = 0.2 v²

5 = 0.2 v²

5 = $\sf\dfrac{2}{10}$v²

Do cross multiplication

5(10) = 2v²

50 = 2v²

v² = 50/2

v² = 25 m/s

v = 5 m/s

So, speed of a body is 5m/s

Know more :-

P.E = mgh

Elastic potential energy = 1/2kx²

Change in K.E = 1/2m(v²-u²)

Law of conservation of mass = Work done by all forces is equal to change in K.E energy

Work done by all external forces = Change in P.E

Answer 2

Answer:

FLW ME

Explanation:

Given :-

Mass = 0.2 kg

Power = 0.5 W

Time = 5 s

u = 0

To find :-

Speed of body after 5 seconds

Formula Implemented:-

P = \sf\dfrac{W}{T}

T

W

Here work done = K.E

P = \sf\dfrac{mv²}{2T}

2T

mv²

Plugging values :-

0.5 = \sf\dfrac{0.2 v²}{10}

10

0.2v²

Do cross multiplication

0.5 (10) = 0.2 v²

5 = 0.2 v²

5 = \sf\dfrac{2}{10}

10

2

v²

Do cross multiplication

5(10) = 2v²

50 = 2v²

v² = 50/2

v² = 25 m/s

v = 5 m/s

So, speed of a body is 5m/s