Question

 A vehicle is moving with a velocity of 9.8 m/s, whose mass is 200 kg. What is the magnitude of force required to stop the vehicle in 10 sec?​

Answer 1

Answer:

Given mass of the vehicle(m)=120 kg and initial velocity u= 90 km/h

u=90km/h=90×(  

18

5

​  

)=25 m/s

As the vehicle is needed to be stopped in 10 seconds the final velocity will be =0 m/s.

We know acceleration=  

time

Final−initialvelocity

​  

 

=  

10

0−25

​  

=−2.5m/s  

2

 

And also Force= mass×acceleration

=120×(−2.5)=−300N  

{The negative sign implies that the force is applied in the direction opposite to that of the direction of motion}  

Explanation:

Answer 2

Answer:

The required magnitude of force is 196 N.

Explanation:

Given :

A vehicle is moving with a velocity of 9.8 m/s, whose mass is 200 kg.

To find :

the magnitude of force required to stop the vehicle in 10 sec

Solution :

From Newton's second law of motion, magnitude of force is given by

⇒ F = ma

where

m denotes the mass of the object

a denotes the acceleration

• Acceleration is defined as the rate of change of velocity.

$\longrightarrow \sf a=\dfrac{v-u}{t}$

where

v denotes the final velocity

u denotes the initial velocity

t denotes the time taken

From the given data,

m = 200 kg

u = 9.8 m/s

t = 10 sec

v = 0 m/s

Substituting the values,

$\sf a=\dfrac{0-9.8}{10} \\\\ \sf a=\dfrac{-9.8}{10} \\\\ \sf a=-0.98 \ m/s^2$

Magnitude of force :

F = ma

F = 200 × (-0.98)

F = -196 N

The negative sign indicates that the force is acted in the direction opposite to that of motion of the vehicle.