Question

A constant force of 2.5 N accelerates a stationary particle of mass 15 g through a displacement of 2.50 m. Find the work done and the average power delivered.

Concept of Physics - 1 , HC VERMA , Chapter " Work and Energy"

Answer 1

Hello Dear.


Given ⇒

Force (F) = 2.5 N.
Mass of the Particle (m) = 15 g.
= 15/1000 kg.
= 0.015 kg.
Displacement covered by the particles(S) = 2.5 m.

For Calculating Work,

∵ Work Done = Force × Displacement
∴ W = F.S
⇒ W = 2.5 × 2.5
∴ Work Done = 6.25 J

Hence, the Work done by the Particles is 6.25 Joules.


For Power,
Work Done = 6.25 J.
Initial Velocity(u) = 0
[∵ Since, the Particles starts from the rest]

Using Newton's second law of Motion,
F = mass × acceleration.
⇒ Acceleration = Force/Mass
∴ a = 2.5/0.015
⇒ a = 500/3 (or 166.67) m/s².

Now, For Calculating the Final Velocity (v), Using the Formula,

v² - u² = 2aS.
⇒ v² - (0)² = 2(500/3)(2.5)
⇒ v² = 2500/3 
⇒ v = √2500/3
∴ v = 50/√3 m/s.

Now, using the Formula for calculating time,

 $a = \frac{v - u}{t}$
⇒ t = (50√3 - 0)/(500/3)
⇒ t = 50/√3 × 3/500
∴ t = 150/500√3 sec.

Now, Power = Work/Time.
⇒ P = 6.25/(150/500√3)
∴ P =  6.25 × 500√3/150
∴ P = 36.08 W.


Hence, Average Power Delivered by the Particle is 36.08 W.


Hope it helps.

Answer 2

Answer:

Given ⇒

Force (F) = 2.5 N.

Mass of the Particle (m) = 15 g.

= 15/1000 kg.

= 0.015 kg.

Displacement covered by the particles(S) = 2.5 m.

For Calculating Work,

∵ Work Done = Force × Displacement

∴ W = F.S

⇒ W = 2.5 × 2.5

∴ Work Done = 6.25 J

Hence, the Work done by the Particles is 6.25 Joules.

For Power,

Work Done = 6.25 J.

Initial Velocity(u) = 0

[∵ Since, the Particles starts from the rest]

Using Newton's second law of Motion,

F = mass × acceleration.

⇒ Acceleration = Force/Mass

∴ a = 2.5/0.015

⇒ a = 500/3 (or 166.67) m/s².

Now, For Calculating the Final Velocity (v), Using the Formula,

v² - u² = 2aS.

⇒ v² - (0)² = 2(500/3)(2.5)

⇒ v² = 2500/3

⇒ v = √2500/3

∴ v = 50/√3 m/s.

Now, using the Formula for calculating time,

⇒ t = (50√3 - 0)/(500/3)

⇒ t = 50/√3 × 3/500

∴ t = 150/500√3 sec.

Now, Power = Work/Time.

⇒ P = 6.25/(150/500√3)

∴ P = 6.25 × 500√3/150

∴ P = 36.08 W.

Hence, Average Power Delivered by the Particle is 36.08 W.