A motorcar is moving with a velocity of $108 km/h$ and it takes $4 s$ to stop after the brakes are applied. Calculate the average force exerted by the brakes on the motorcar if it's mass along with the passengers is $1000 kg$ .

4500 N

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5500 N

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7500 N

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2500 N

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Solution

The correct option is C $7500 N$
Average force,

$F = \frac{Δ p}{Δ t} = \frac{p_{f} - p_{i}}{Δ t} = \frac{m (v_{f} - v_{i})}{Δ t}$

$\Rightarrow F = \frac{1000 (0 - 108 \times 5 / 18)}{4} = - 7500 N$

Negative sign shows that brake force acts opposite to the initial velocity.

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A motorcar is moving with a velocity of 108 km/h and it takes 4 s to stop after the brakes are applied. Calculate the average force exerted by the brakes on the motorcar if it's mass along with the passengers is 1000 kg.

The correct option is C 7500 NAverage force, F=ΔpΔt=pf−piΔt=m(vf−vi)Δt ⇒F=1000(0−108×5/18)4=−7500 N Negative sign shows that brake force acts opposite to the initial velocity.

A motorcar is moving with a velocity of $108 km/h$ and it takes $4 s$ to stop after the brakes are applied. Calculate the average force exerted by the brakes on the motorcar if it's mass along with the passengers is $1000 kg$ .

The correct option is C $7500 N$
Average force,

$F = \frac{Δ p}{Δ t} = \frac{p_{f} - p_{i}}{Δ t} = \frac{m (v_{f} - v_{i})}{Δ t}$

$\Rightarrow F = \frac{1000 (0 - 108 \times 5 / 18)}{4} = - 7500 N$

Negative sign shows that brake force acts opposite to the initial velocity.