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A car of mass 480 kg moving at a speed of 54 km h-1 is stopped by applying brakes in 10 s. Calculate the force applied by the brakes.

Laws of Motion

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Answer

We know that,

acceleration = vut\dfrac{v - u}{t}

Given,

m = 480 kg

u = 54 km h-1

converting km h-1 to m s-1

54km h1=54×(1000m60×60s)54km h1=54×(10m36s)54km h1=540m36s54km h1=15m s154 \text {km h}^{-1} = 54 \times (\dfrac{1000 \text {m}}{60 \times 60 \text {s}}) \\[0.5em] 54 \text {km h}^{-1} = 54 \times (\dfrac{10 \text {m}}{36 \text {s}}) \\[0.5em] 54 \text {km h}^{-1} = \dfrac{540 \text {m}}{36 \text {s}} \\[0.5em] 54 \text {km h}^{-1} = 15 {\text {m s}^{-1}} \\[0.5em]

Hence, u = 15 m s-1

v = 0

t = 10 s

Substituting the values in the formula above we get,

a=01510a=1.5m s2a = \dfrac{0 - 15}{10} \\[0.5em] \Rightarrow a = -1.5 \text{m s}^{-2} \\[0.5em]

Hence, a = - 1.5 m s -2

The negative sign shows that it is retardation.

Now,

Force (f) = mass (m) x acceleration (a)

Substituting the values in the formula above we get,

F=480×(1.5)F=720NF = 480 \times (1.5) \\[0.5em] \Rightarrow F = 720 N \\[0.5em]

Hence,

Magnitude of force applied by the brakes = 720 N

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