Question

A constant retarding force of 50N is applied to a body of mass 20kg moving initially with a speed of 15m\s . How long does the body take to stop?​

Answer 1

Given :-

• Retarding Force = -50 N
• Mass = 20 kg
• Initial Speed = 15 m/s
• Final Velocity = 0 m/s

To Find :-

• How long does the body take to stop?

Answer :-

• Time taken by the body to stop is 6 seconds.

Explaination :-

By using Newton's second law of motion we get :-

→ Force = mass × Acceleration

→ -50 = 20 × Acceleration

→ Acceleration = -50/20

→ Acceleration = -5/2

→ Acceleration = -2.5 m/s²

• Calculating time taken by the body to stop :

By using first equation of motion :

→ v = u + at

Where,

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

Now, substitute the given values in above equation :

→ 0 = 15 + (-2.5) t

→ 2.5t = 15

→ t = 15/2.5

→ t = 6 seconds

Therefore,time taken by the body to stop is 6 seconds.

Answer 2

$\LARGE{\bf{\underline{\underline{GIVEN:-}}}}$

• Force (f) = –50N. [It is negative because force is in opposite direction]
• Mass (m) = 20 kg.
• Initial velocity (u) = 15ms⁻¹.
• Final velocity (v) = 0 [Because the object stops due to the opposite force].

$\LARGE{\bf{\underline{\underline{TO \ FIND:-}}}}$

• The time taken for the body to stop (t).

$\LARGE{\bf{\underline{\underline{SOLUTION:-}}}}$

First, let us find out the acceleration.

$\mapsto \sf{\green{F=ma}}$

Where:

• F is force.
• m is mass.
• a is acceleration.

Substitute the values.

$\sf \to -50 = 20 \times a$

$\to \sf a=-\dfrac{50}{20}$

$\leadsto \sf{\red{a=-2.5 ms^{-1}}}$

Recall the three equations of motion:

$\boxed{\substack{\displaystyle \sf v=u+at \\\\ \displaystyle \sf s=ut+\dfrac{1}{2}at^2 \\\\ \displaystyle \sf 2as=v^2-u^2}}$

It is apt to use the first equation because we are given v, u and a and we can easily find out t.

$\mapsto \sf{\green{v=u+at}}$

Where:

• v is final velocity.
• u is initial velocity.
• a is acceleration.
• t is time.

Substitute the values,

$\sf \to 0=15 + (-2.5) \times t$

$\sf \to t=\dfrac{-15}{-2.5}$

$\leadsto \sf{\red{t=6 sec.}}$

TIME IS 6 SECONDS.