Question

200 kg car traveling at 10 m/s hits a tree and is brought to rest in 0.10 s. What is the magnitude of the average force acting on the car to bring it to rest?

Answer 1

Given:-

• Mass of car = 200kg

• Initial Velocity = 10m/s

• Time Taken = 0.10s.

• Final Velocity = 0m/s (as it comes to rest).

To Find:-

• The Average force exerted by tree.

FORMULAE USED:-

• v = u + at

• F = ma.

Where,

v = Final Velocity

u = Initial Velocity

a = Acceleration

t = Time.

Now,

v = u + at.

(0) = 10 + a × 0.10

-10 = 0.10a

-10 = 0.10/100

-10 × 100 = 10

-1000 = 10a

a = -1000/10

a = -100m/s²

Hence, The Retardation is 100m/s² .

Negative sign is implies that Acceleration is acting on the opposite direction.

Therefore,

F = ma

F = 200 × -100

F = -20000N.

Hence, The average force acting on the car to bring it to rest is -2000N

Negative sign implies that force is acting in Opposite Direction.

Answer 2

✰ Given :-

• Mass (m) = 200kg
• Initial Velocity (u) = 10m/s
• Time (t) = 0.10sec
• Final Velocity (v) = 0m/s

✰ To Find :-

• Force exerted by car

✰ Solution :-

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We know Equation Of Kinematics,

$✓ \: v = u + at$

Placing Values,

$⟶(0) = 10 + a \times 0.10$

$⟶ - 10 = a \times 0.10$

$⟶ - 10 = 0.10a$

$⟶ - 10 = \frac{\cancel{0.10}}{\cancel{100}}$

$⟶ - 10 \times 100 = 10a$

$⟶ - 1000 = 10a$

$⟶a = \frac{\cancel{-1000}}{\cancel{10}}$

$⟶a = - 100 {m</strong><strong>/</strong><strong>s}^{2}$

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Now we know Formula for Force,

$✓ \: f = ma$

Placing Values,

$⟶f = 200 \times - 100$

$⟶f = - 20000</strong><strong>N</strong><strong>$

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⛬ Force Exerted by tree is -20000N (Newton's)