A vehicle of 100 kg is moving with a velocity of 5m/sec. Top stop it in 1/10sec.The required force in opposite directions is

Answers 1

Given,

Mass of the vehicle = 100 kg.

Initial velocity (u) = 5 m/sec

Final Velocity (v) = 0 m/sec

Time taken to stop it (t) = 1/10 sec

To Find,

Given the initial conditions, the required fore in the opposite direction to stop the car.

Solution,

We can solve this numerical by the following process,

We know F=m*a

where F= The amount of applied force

m= mass of the vehicle

a= acceleration of the vehicle

(Here all of the physical quantities are in SI units, So we do not have to do any unit conversion.)

Here, [tex]a=\frac{v-u}{t}\\ a=\frac{0-5}{\frac{1}{10} }\\a=(-5)*10\\a=-50[/tex]

Hence, acceleration is -50 [tex]m/s^2[/tex]

Therefore, Required force, F=m*a

F = (100*(-50))

F = -5000 N

The negative sign signifies that the required force is applied in the opposite direction.

Hence, the required force is -5000 N.