Question

A car of mass 1800 kg moving with the speed of 10 m/s is brought to rest after covering a distance of 50 m . Calculate the force applied to the car

Answer 1

Using third law of 1d kinematics,

v^2-u^2=2as

Given, s=50m; v=0; u=10m/s

Then , v^2-u^2=2as

0-(10)^2=2a(50)

a= -1m/s

Force=mass×acceleration, (or) Force=mass×deceleration

Therefore, deceleration= -1m/s^2

Deceleration= -(acceleration)

Force=mass×deceleration

=1800×1

=1800N

Force will be acting in opposite direction to displacement of the car

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Answer 2

Answer:

Explanation:

Solution,

Here, we have

Mass, m = 1800 kg

Initial velocity, u = 10 m/s

Final velocity, v = 0 (As car brought to rest)

Distance covered, s = 50 m.

To Find,

Force applied, F = ??

At, 1st we have to find the acceleration, a,

According to the 1st equation of motion,

We know that,

v² - u² = 2as

So, putting all the values, we get

⇒ v² - u² = 2as

⇒ (0)² - (10)² = 2 × a × 50

⇒ 100 = 100 a

⇒ 100/100 = a

⇒ a = 1 m/s²

Here, the acceleration of the car is 1 m/s².

Now, we will find Force applied, F,

According to the 2nd equation of Newton,

We know that,

F = ma

So, putting all the values again, we get

⇒ F = ma

⇒ F = 1800 × 1

⇒ F = 1800 N

Hence, the force applied to the car is 1800 N.