First slide

Work

Question

A car moving with a speed of 50km/h can be stopped by brakes after at least 6m. If the same car is moving at a speed of 100km/h, the minimum stopping distance is

Moderate

A

6m
B

2m
C

18m
D

24m

Solution

Using work-energy theorem, $W_{t o t a l} = Δ K$
i.e, work done by stopping forces should be equal to change in the kinetic energy of the car.
$- F_{s t o p p i n g} x_{s t o p p i n g}$ distance= $(k_{f i n a l} - k_{i n i t i a l})$
$- F_{s t o p p i n g}$ x S= $(0 - \frac{1}{2} m {v^{2}}_{i n i t i a l})$
$\Rightarrow - F_{s t o p p i n g}$ x S = $\frac{1}{2} m {v^{2}}_{i n i t i a l}$
As stopping force is same for both the cars, hence
$S α v^{2} {}_{i n i t i a l}$ =>Hence , $\frac{S_{1}}{S_{2}} = \frac{{({v^{2}}_{i n i t i a l})}_{1}}{{({v^{2}}_{i n i t i a l})}_{2}}$
Or $\frac{6}{S_{2}} = \frac{{({v^{2}}_{i n i t i a l})}_{1}}{{({v^{2}}_{i n i t i a l})}_{2}} = {(\frac{50}{100})}^{2} \Rightarrow S_{2} = 24 m$

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