Question

A bus weighing 50 tonnes (1 tonne=
= 1000kg) is moving with a velocity ,
of 60 km/h. calculate the force
required to stop it in 10 sec.​

Answer 1

Answer:-

F = - 83,300 N

Given :-

v = 0m/s

u = 60 km/hr

= 60 × 5/18

= 16.66 m/s

t = 10 s

m = 50 tonnes

m = 50 × 1000

m = 50000 kg

To find :-

The force required to stop the bus.

Solution:-

The force required to stop will be given by using Newton 2 nd law of motion:-

$\huge \boxed {F = m \times \dfrac{v-u}{t}}$

→$F = 50,000 \times \dfrac{0-16.66}{10}$

→$F = 50,000 \times \dfrac{-16.66}{10}$

→ $F = 50,000 \times -1.666$

→$F = -83, 300 N$

hence,

The frictional force applied to stop the bus will be -83, 300 N .

Answer 2

Answer:

The force applied to stop the bus is 83, 300 N

Explanation:

Correct Question:

A bus weighing 50 tonnes (1 tone =  000kg) is moving with a velocity , of 60 km/h. calculate the force  required to stop it in 10 sec.​

Solution:

To Find:

The force required to stop the bus.

-------------------------

Method:

Given that,

v = 0m/s  

u = 60 km/hr  = 60 × 5/18   =16.66 m/s

t = 10s

m =  50 × 1000  = 50000 kg

We know that,

Newton Newton 2nd law of motion:

$\sf \bigstar F = m\times \dfrac{v-u}{t}\bigstar$

Put the given values:

$\sf => F = 50,000 \times \dfrac{0-16.66}{10}\\\\\ =>F = 50,000 \times \dfrac{16.66}{10}\\\\\ =>F = 50,000 \times 1.666 = 83,300\\\\\bigstar Hence,\\\ The\; force\; applied\; to\; stop\; the\; bus\; is \;83,300N\bigstar$