The constant force resisting the motion of a car of mass $1500 \; kg$ is equal to one fifteenth of its weight. When travelling at $48 \; km/h$, the car is brought to rest in a distance of $50 \; m$ by applying the brakes, find the additional retarding force due to the brakes (assumed constant) and heat developed in the brakes.

Given,

Mass of car (m) = 1500 kg

Frictional force ($F_c)$ = $\frac{1}{15}$$\;m.g$

                                      = $\frac{1}{15}$$\;*\;1500\;*\;10 = 1000\;N$

Initial velocity (u) = 48 km/hr = $\frac{48\;*\;1000}{60\;*\;60}$ $= 13.33\;m/s$

Distance covered (s) = 50 m

Final velocity (v) = 0

Additional force ($F_a = \;?$)

Heat developed in the engine = ?

To calculate, $a = ?$ We have,

$v^2 = u^2 \;+\;2.a.s$

$0^2 = (13.33)^2 + 2\;*\;a\;*\;50$

⇒ a = $-$ $\frac{(13.33)^2}{2\;*\;50}$$ = -\;1.78\;m/s$

The retarding force due to friction is,

$F_r = m\;a = -\;1.78\;*\;1500 = 2666.67\;N$

But frictional force ($F_c = 1000\;N$)

So, Additional force ($F_a$) = $F_r - F_c = 2666.67 - 1000 = 1666.67\;N$

Now heat developed = ?

Heat developed = Work done

i.e. $W = F_a\;*\;s = 1666.67\;*\;50 = 83333.33\;J$

∴ Additional retarding force due to brake is $1666.67\;N$

∴ Heat developed in the engine is $83333.33\;J$

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