Question

a body of mass2 kg moving with a velocity of 10ms—¹ experience a force of 50 N which act on it for 0.2s . calculate the velocity of body after the force ceases to act.​

Answer 1

Given :

▪ Mass of body = 2kg

▪ Initial velocity = 10m/s

▪ Force acts on the body = 50N

▪ Time interval = 0.2s

To Find :

▪ Final velocity of body after given interval of time.

Solution :

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

→ Force can be measured from Newton's second law of motion...

☞ F = ma = m×(Δv/Δt)

• F denotes force
• m denotes mass
• a denotes acceleration
• Δv denotes change in velocity
• Δt denotes time interval

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✏ F = 50N

✏ m = 2kg

✏ Δt = 0.2s

✏ Δv = (v - u) = (v - 10)

• v denotes final velocity
• u denotes initial velocity

✒ F = m×(Δv/Δt)

✒ 50 = 2×[(v - 10)/0.2]

✒ 50 = 10×(v - 10)

✒ 5 = v - 10

✒ v = 15mps

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Additional information :

• Force is a polar vector as it has a point of application.
• Force can be classified as positive and negative.
• A positive force represents repulsion and a negative force represents attraction

Answer 2

Given :

• Mass of Body (m) = 2 kg
• Initial velocity (u) = 10 m/s
• Force (F) = 50 N
• Time (t) = 0.2 s

To Find :

• Velocity of body after the force acts, or final velocity

Solution :

As we know that,

$\implies \sf{F \: = \: ma} \\ \\ \implies \sf{F \: = \: \dfrac{m(v \: - \: u)}{t}} \\ \\ \implies \sf{F \: = \: \dfrac{2(u \: - \: 10)}{0.2}} \\ \\ \implies \sf{50 \: = \: 10(u \: - \: 10)} \\ \\ \implies \sf{u \: - \: 10 \: = \: \dfrac{50}{10}} \\ \\ \implies \sf{u \: - \: 10 \: = \: 5} \\ \\ \implies \sf{u \: = \: 5 \: + 10} \\ \\ \implies \sf{u \: = \: 15}$

$\therefore$ Velocity after action of force is 15 m/s