Question

The car of mass 200 kg moving at 36 km/h is brought to rest after it covered a distance of 10 m. Find the retarding force acting on the car ?

Answer 1

Answer:

F = 1000 N

Step by step explanation:

It is given that:

• Mass 'm' = 200 kg
• Initial velocity $v_i$= 36 km/h = $\frac{36*1000}{60*60}=10\:m/s$
• Distance 'd' = 10 m
• Final velocity $v_f$=0

In order to find the force, we first need to calculate the acceleration.

Using the Kinematics law:

$v_f^2-v_i^2=2ad\\0-100=2a*10\\a=-5\:m/s^2$

Therefore the deceleration = $5\:m/s^2$.

F = mass*deceleration

F = 200*5

F = 1000 N

Answer 2

Mass = m = 200kg

intial velocity = u = 36km/hr = 36000m/3600s = 10m/s

k

final velocity = v = 0 km/hr = 0m/s

distance = s = 10m

acceleration = a = ?

Force = F = ?

____________________

v²-u² = 2as

(0)²-(10)² = 2(a)(10)

0-100 = 20a

-100/20 = a

-5 m/s² = a

___________________

F = ma

F = 200kg × -5m/s²

F = -1000kgm/s²

F = -1000N

F = -10³N

F = -1 KN

(here minus sign shows that force acting apposite of objects motion )

so force acting on the car is -1 kilo newton. or -1000 newton

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