Question

①• A force is applied on a body of mass 20 kg moving with a velocity of 40 m/s. The body attains a velocity of
50 m/s in 2 s. Calculate the work done on the body.

②• A girl of mass 35 kg climbs up from the first floor of a building at a height 4 m above the ground to the third floor at a height 12 m above the ground. What will be the increase in her gravitational potential energy? (Take g = 10 m/s²)

❌❗No SpaMinG❗❌
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Answer 1

1st answer:

Given that,

• mass = m = 20 kg
• initial velocity = u = 40 m/s
• Final velocity = v = 50 m/s
• Time taken = t = 2 s

We know that,

• a = (v - u)/t

Putting the values given in the question,

a = (50 - 40)/2

→ a = 10/2

→ a = 5 m/s²

We know that,

• s = ut + (1/2)at²

Putting the values given in the question,

→ s = (40 * 2) + (1/2)(5 * 2²)

→ s = 80 + 10

→ s = 90 m

We know that,

• F = ma

Putting the values given in the question,

→ F = 20 * 5

→ F = 100 N

We know that,

• W = F s cos∅

where W is work done, F is force, s is displacement and ∅ is the angle between force and displacement.

Considering the force and displacement are in same direction, ∅ = 0°.

→ W = 100 * 90 * cos0°

→ W = 9000 J = 9 kJ Ans.

2nd answer:

Given that,

• mass = m = 35 kg
• initial height = h₁ = 4 m
• Final height = h₂ = 12 m

We can consider any point to be a reference point and say that the gravitational potential energy (GPE) at that point will be 0. The GPE at any other point will be calculated in reference to the reference point.

We know that,

• GPE = mgh

where m is the mass, g is the acceleration due to gravity, and h is the displacement between where the object is and the reference point.

Considering the ground level to be reference point,

• Initial GPE = mgh₁
• Final GPE = mgh₂

Then, change in GPE = mgh₂ - mgh₁

Putting the values given in the question,

→ ∆GPE = mg(h₂ - h₁)

→ ∆GPE = 35 * 10 * (12 - 4)

→ ∆GPE = 35 * 10 * 8

→ ∆GPE = 2800 J = 2.8 kJ Ans.