Question

How much force is required to stop a bus (10second) of mass 5000kg moving with a speed of 72km/h?


please solve this problem in details.....

Answer 1

How much force is required to stop a bus (in 10 seconds) of mass of 5000 kg moving at a speed of 72km/h?

Initial velocity u of the bus = 72 km/h= 20 m/s

Finalvelocity v of the bus when it is stopped= 0 m/s

Time t in which the bus is stopped= 10 s

The acceleration 0n the bus= ( v - u)/t = ( 0m/s - 20 m/s)/ 10s= - 2 m/s² ( minus sign implies deceleration)

Force used to stop the bus = mass of the bus in kg× deceleration = 5000 kg× - 2 m/s² = - 10,000 N.

The minus sign implies force is supplied by the decelerating agency.

Answer 2

Explanation:

by using 1st equation of motion

when u=72kmph

5/18×72=20m/s

v=0

mass=5000kg

t=10sec

v=u+at

0=20m/s+a×10sec

-20/10=a

-2m/s^2=a

F=ma

F=5000kg ×(-2m/s)

F=-10000N