Question

The work done by a force on body of mass 5 kg to accelerate it in the direction of force from rest to 20 m/s² in 10 second is

• A. 2x10³ J
• B. 4x10³ J
• C. 10^-3 J
• D. 10³ J

Hint:

Acceleration, Force, and Distance
Acceleration (a): The acceleration that causes a speed increase from 0 to 20 m/s in 10 seconds is calculated as:
a = v / t = 20 m/s / 10 s = 2 m/s²
Force (F): The force acting on the body, given mass (m) of 100 kg, is:
F = m * a = 100 N * 2 m/s² = 200 N
Distance (s): The distance traveled in the given time (10 seconds) is:
s = 1/2 * a * t² = 1/2 * 2 m/s² * (10 s)² = 100 m
Therefore,

• The acceleration is 2 m/s².
• The force acting on the body is 200 N.
• The distance traveled in 10 seconds is 100 meters.

Answer 1

The Correct Option is D

acceleration = \(\frac {final \ velocity - initial \ velocity}{time}\)
acceleration = \(\frac {20 -0}{10}\)= 2 m/s². 
Force = mass × acceleration. 
Given that the mass is 5 kg and the acceleration is 2 m/s²: 
Force = 5 kg × 2 m/s² = 10 N. 
Work = Force × Displacement × cos(θ). 
Displacement = \(\frac {1}{2}\) × acceleration × time²
Displacement = \(\frac {1}{2}\) × 2 m/s² × (10 s)² = 100 m. 
Work = 10 N × 100 m × 1 = 1000 J. 
Therefore, the work done by the force is 1000 Joules (J). 
The correct option is (D) 10³ J.