Question

A body of mass 10 kg is moving with a velocity of 30 m/s. When brakes are applied, it stops after travelling a distance of 15 metres. Find retardation and retarding force​

Answer 1

Answer:

-300 N

Explanation:

Given

Mass = 10 kg

u = 30 m/s

v = 0 m/s

s = 15 m

Force = ma

= m(v²-u²)/2s

= 10 {(0²)-(30²)/2(15)

= 10 (-900/30)

= 10 (-30)

= -300 N

Hope it helps!!

Answer 2

Given :

• Mass (m) = 10kg
• Initial Velocity (u) = 30m/s
• Final Velocity (v) = 0m/s
• Distance Travelled (s) = 15 metres

To Find :

• Retardation
• Retarding Force

Solution :

If the speed is decreasing with time then accerlation is negative. The negative accerlation is called retardation.

Retardation = - ( acceleration )

$\implies \displaystyle{\sf v^2 = u^2+2as}$

$\implies \displaystyle{\sf 2as = v^2-u^2}$

$\implies \displaystyle{\sf 2a = \frac{v^2-u^2}{s}}$

$\implies \displaystyle{\sf a = \frac{v^2-u^2}{2s}}$

Now substitute the values in formula

$\Rightarrow \displaystyle{\sf a = \frac{v^2-u^2}{2s}}$

$\Rightarrow\displaystyle{\sf a = \frac{0^2-30^2}{2(15)}}$

$\Rightarrow\displaystyle{\sf a = \frac{0-900}{30}}$

$\Rightarrow\displaystyle{\sf a = -30}$

$\Rightarrow\sf Acceleration = -30 m/s^2$

$\sf\boxed{ \sf Retardation = 30 m/s^2}$

Force = Mass × Acceleration

$\Rightarrow 10 \times -30$

$\Rightarrow -300$

Negative sign in Acceleration force is Positive sign in Retardation. Hence,

$\boxed{\sf Retarding\: force = 300N}$

Required Answer :

Retardation of the man is $\sf \underline{30 m/s^2}$

Retarding Force of the man is $\underline{\sf 300N}$