A 1200 kg car is braked to a stop from 100 km/h. If 60% of this energy appears in the steel brake drums, whose total mass is 10 kg its temperature will rise by $(s_{steel} = 450 J / k g K)$

{31.2}^{\circ} C

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{61.7}^{\circ} C

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{92.1}^{\circ} C

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{42.1}^{\circ} C

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Solution

The correct option is B ${61.7}^{\circ} C$
$\begin{matrix} A c c o r d i n g t o q u e s t i o n . . . . . . . . . . . . . H e r e, m = 1200 k g, b r e a k d r u m (m) = 10 k g u = 100 k m / h = 27.77 m / s S_{S t e e l} = 450 J / K g K, V = 0 m / s s o, E n e r g y r e l e a s e d d u r i n g b r e a k i n g = c h a n g e i n k i n e t i c e n e r g y = \frac{1}{2} m v^{2} - \frac{1}{2} m u^{2} = \frac{1}{2} (1200) (0)^{2} - \frac{1}{2} (1200) (27.77)^{2} = 0 - 462937.0374 J = 462937.0374 J N o w, 60 p e r e c e n t o f t h i s e n e r g y a p p e a r s = 0.6 \times 462937.0374 J = 277762.225 F o r c h a n g e i n t e m p e r a t u r e : m s Δ \frac{1}{2} t = 277762.225 \Rightarrow Δ t = \frac{277762.225}{m s} = \frac{277762.225}{10 \times 450} ∴ Δ t = {61.7}^{0} C s o t h e c o r r e c t o p t i o n i s B . \end{matrix}$

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A 1200 kg car is braked to a stop from 100 km/h. If 60% of this energy appears in the steel brake drums, whose total mass is 10 kg its temperature will rise by (s steel =450J/kgK)

A 1200 kg car is braked to a stop from 100 km/h. If 60% of this energy appears in the steel brake drums, whose total mass is 10 kg its temperature will rise by $(s_{steel} = 450 J / k g K)$