Question

A force acts upon body of mass 20 Kg, initially at rest, for 6 seconds, after which the force ceases. Now the body describes 60 m in the next 5 Seconds. Find the magnitude of the force ​

Answer 1

Given,

Initial velocity of the body = 0 m$s^{-1}$

Final velocity  of the body = $\frac{distance}{time}$

                      = $\frac{60m}{5s}$

                      = $12ms^{-1}$

Now We know,

Acceleration  = $\frac{FinalVelocity-InitialVelocity}{timetaken}$

                              = $\frac{v-u}{t}$

                               = $\frac{12ms^{-1}}{6s}\\\\=2ms^{-2}$

By Newton's second law of motion,

F = ma

= $20kg*2ms^{-2} \\\\= 40N$

Thus the magnitude of Force is 40N.

Answer 2

Answer:

The magnitude of Force is 40N.

Explanation:

Given :

Initial velocity of the body = 0 ms⁻¹

To Find :

The magnitude of the force.

Solution :

• First Step :
• Final velocity of the body.

We know that :

$\bigstar\;\boxed{\sf \dfrac{Distance}{Time}}}$

So,

$\sf\\\\\implies \dfrac{60m}{5s}\\ \\ \\ \implies 12ms^{-1}$

• Second Step :
• Find It's Accelration.

We know that :

$\bigstar\;\boxed{\sf Acceleration = \dfrac{Final\;Velocity-Initial\;Velocity}{Time\;taken}}}$

So,

$\sf\\ \\\implies \dfrac{12ms^{-1}}{6s}\\ \\ \\\implies 2ms^{-2}$

Now,

We have :

Initial velocity of the body = 0 ms⁻¹

Final Velocity of the body = 12 ms⁻¹

Accelation = 2 ms⁻²

• Third Step :
• Use Newton's second law of motion.

Now,

We know that :

Newton's second law of motion :

$\bigstar\;\boxed{\sf F = ma}}$

Now,

$\sf\\ \\\implies 20kg\times2ms^{-2}\\ \\ \\\implies 40N$

∴ The magnitude of Force is 40N.