A force of 6 N acts horizontally on a stationary mass of 2 kg for 4 s.

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

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

Answer»

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

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