Question

Q13 A constant force acts on an object of mass 4 kg for a duration of 3 seconds. It increases the object's velocity from 3 m/s to 9 m/s. Find the magnitude of the applied force.
​

Answer 1

Given :-

• A constant force acts on an object of mass 4 kg for a duration of 3 seconds.
• It increases the object's velocity from 3 m/s to 9 m/s.

To Find :-

• What is the magnitude of the applied force ?

Solution :-

First, we have to find the acceleration :-

Given :

• Initial Velocity (u) = 3 m/s
• Final Velocity (v) = 9 m/s
• Time Taken (t) = 3 seconds

According to the question by using the formula we get,

$\implies \sf\boxed{\bold{v =\: u + at}}$

where,

• v = Final Velocity
• u = Initial Velocity
• a = Acceleration
• t = Time Taken

So, by putting those values we get,

$\implies \sf 9 =\: 3 + a(3)$

$\implies \sf 9 - 3 =\: 3a$

$\implies \sf 6 =\: 3a$

$\implies \sf \dfrac{6}{3} =\: a$

$\implies \sf 2 =\: a$

$\implies \sf\bold{a =\: 2\: m/s^2}$

Hence, the acceleration is 2 m/s² .

Now, we have to find the magnitude of the applied force :

Given :

• Mass (m) = 4 kg
• Acceleration (a) = 2 m/s²

According to the question by using the formula we get,

$\implies \sf\boxed{\bold{F =\: ma}}$

$\implies \sf\bold{Force =\: Mass \times Acceleration}$

$\implies \sf Force =\: 4 \times 2$

$\implies \sf\bold{\underline{Force =\: 8\: N}}$

$\therefore$ The magnitude of the applied force is 8 N .

Answer 2

Answer:

Magnitude of the applied force is 8N

Explanation:

Mass, m = 4kg

Time, t = 3s

Initial velocity, u = 3m/s

Final velocity, v = 9m/s

Magnitude = ?

To find acceleration, a

v = u + at

9 = 3 + a × (3)

3a = 9 - 3

3a = 6

a = 2m/s²

To find magnitude,

F = ma

F = 4kg × 2m/s²

F = 8N

Hence the magnitude of the applied force is 8N.