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Brut [27]
3 years ago
15

A 1500 kg car drives around a flat, 50 m diameter track, starting from rest. The drive wheels supply a small but steady 525 N fo

rce in a forward direction. The coefficient of static friction between the tires and the road is 0.90. How many revolutions of the track have been made when the car slides out?
Physics
1 answer:
krok68 [10]3 years ago
5 0

Answer:

No. of revolutions before the car slides off is 2

Solution:

As per the question:

Mass of the car, m = 1500 kg

Diameter of the track, d = 50 m

Force, F = 525 N

Coefficient of static friction, \mu_{s} = 0.90

Now,

The acceleration in the tangential direction is given by:

F = ma_{T}

525 = 1500\times a_{T}

a_{T} = 0.35\ m/s^{2}

Here, the centripetal force is given by the friction force:

\frac{mv^{2}}{R} = \mu mg

Thus

v = \sqrt{\mu gR} = \sqrt{0.90\times 9.8\times 25} = 14.849\ m/s

Time taken by the car is given by:

v = v' + at

v' = initial velocity = 0

Thus

t = \frac{v}{a} = \frac{14.849}{0.35} = 42.426\ s

Total Distance covered, d is given by:

v^{2} = v'^{2} + 2ad

14.849^{2} = 0^{2} + 2\times 0.35d

d = 341.99 m

Distance covered in 1 revolution is the circumference of the circle, d' = 2\pi R

Now, the no. of revolutions is given by:

n = \frac{d}{d'} = \frac{314.99}{2\pi \times 25} = 2

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3 years ago
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