Question

A car of 900 kg mass is moving at the velocity of 60 km/hr. It is
brought into rest at 50 metre distance by applying brake. Now,
calculate the force required to stop the car. [Ans: 2500 N]

IN COPY PLEASE

Answer 1

Given:

Mass of car =900 kg

Initial velocity = 60km/h=16.667 m/s

Final velocity = 0

Distance travelled = 50 m

To Find:

Force applied

Formula :

F = m× a

where,

F is force applied

m is mass of the object

a is acceleration produced

Solution:

To find acceleration -

V^2 = U^2 + 2as

=> (0)^2 = (16.667)^2 + 2×a × 50

=> 0 = 277.78 + 100a

=> -277.78 = 100a

=> a = -277.78/ 100

=> a = -2.7778 m/s^2

Now,

F = m×a

F = 900 × 2.7778

F = -2500.2 N

in approx = -2500N

opposing force = 2500N

Answer 2

Answer:

2500 N

Explanation:

Given:

• Mass of the car = m = 900 kg

• Initial velocity of the car = u = 60 km/hr = $60 \times \frac{5}{18} =\frac{50}{3} \ m/s$

• Final velocity = v = 0 m/s
• Distance travelled = s = 50 metres

To find:

• Force required to stop the car

Using third equation of motion:

$0^{2} -\frac{50}{3 }^{2} = 2 \times a \times 50$

$\frac{-2500}{9} =100a$

a=$\frac{-2500}{9} \times \frac{1}{100}$

a=$\frac{-25}{9} m/s^{2}$

Using first equation of motion:

0=$\frac{50}{3} -\frac{25}{9} \times t$

$-\frac{50}{3} = \frac{-25}{9}t$

t = $\frac{-50}{3} \times \frac{-9}{25}$

t = 6 seconds

Force = $\frac{Mass(final \ velocity - initial \ velocity)}{Time}$

Force =$\frac{900(0-\frac{50}{3}) }{6}$

Force =$\frac{300 \times -50}{6}$

Force =$\frac{-15000}{6}$

Force = -2500 N in direction of car

2500N is required to stop the car