Question

A force of 6 N acts horizontally on a stationary mass of 2 kg for 4 s. The K.E in joule is
A) 12
B) 75
C) 144
D) 48
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Answer 1

Given :

▪ Initial velocity = zero

▪ Applied force = 6N

▪ Mass of body = 2kg

▪ Time interval = 4s

To Find :

▪ KE of body

Concept :

☞ This question is completely based on the concept of newton's second law of motion.

☞ As per this law, force is defined as the rate of change in linear momentum.

Mathematically,

✴ F = ΔP/Δt

☞ Momentum is defined as the product of mass and velocity.

✴ P = m × v

☞ Formula of kinetic energy is given by

✴ K = 1/2×mv²

Calculation :

⚽ Final velocity :

→ F = ΔP/Δt

→ F = mΔv/Δt

→ F = m(v-u)/Δt

→ 6 = 2(v-0)/4

→ v = 24/2

→ v = 12m/s

⚽ Kinetic energy :

→ K = 1/2×mv²

→ K = 1/2×2×(12)²

→ K = 144J

Answer 2

YOUR QUESTION :

A force of 6 N acts horizontally on a stationary mass of 2 kg for 4 s. The K.E in joule is

A) 12

B) 75

C) 144

D) 48

YOUR ANSWER :

Given :

• Initial velocity = zero.
• Applied force = 6N.
• Mass of body = 2kg.
• Time interval = 4s.

To find :

• KE of body.

Concept :

✔As per this law, force is defined as the rate of change in linear momentum.

Mathematically,

✔F = ∆P/∆t.

✔P = m × v.

✔K = 1/2 × mv².

Calculation :

✔Final velocity

=> F = ∆P/∆t

=> F = m∆v/∆t

=> F = m(v-u)/∆t

=> 6 = 2(v-0)/4

=> v = 24/2

=> v = 12m/s

✔Kinetic energy

=> K = 1/2×mv²

=> K = 1/2×2×(12)²

=> K = 144J