Question

A man moves on a straight horizontal road with a block of mass 2 kg in his hand, If he covers a distance of 40 m with an acceleration of 0.5 m/s², find the work done by the man on the block during the motion.
Concept of Physics - 1 , HC VERMA , Chapter " Work and Energy"

Answer 1

Hello Dear.

Given conditions :-
Mass of a block (m) = 2 kg.
acceleration (a) = 0.5 m/s²
distance (s) = 40 m.
Now velocity v after he cover a distance 40 m . So applying the formula,
v² = u² + 2as
v² = 0² + 2 × 0.5 × 40
v² = 40 

Using the work energy theorem , work done in moving body is equal to the change in kinetic energy
∴ The work done = Change in Kinetic Energy .
  work done= 1/2 mv² - 1/2 mu²
= 1/2 m ( v²- u²)
= $\frac{1}{2}$  × 2 ( 40 -0²)
= 40 J.

The work done is 40 J.


Hope it Helps.

Answer 2

Answer:

Work Done = Force × Displacement

We know that,

⇒ Force = Mass × Acceleration

⇒ Work Done = Mass × Acceleration × Displacement

According to the question,

• Mass = 2 kg
• Acceleration = 0.5 m/s²
• Displacement = 40 m

Substituting in the formula we get,

⇒ Work Done = 2 × 0.5 × 40

⇒ Work Done = 80 × 0.5

⇒ Work Done = 40 J

Hence the work done by the man in carrying the block is 40 J.

Hope it helped !!

Thanks !!