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`        A car is travelling at the rate of 54kmph . Suddenly the brakes are applied causing tyres to skid .How far will the car travel  before coming to rest if coefficient of friction is 0.2 and g is equal to 9.8`
2 years ago

## Answers : (1)

Dinesh Nayak
62 Points
```							Frictional force = u(coefficient of friction) * F(Normal force)
We know,
Energy = Force * Displacement
therefore, Frictional energy = u * F * S(Displacement)
Similarly, we know, kinetic energy = ½ [m * v^2]
Since energy is same throughout
½ [m * v^2] = u * F * S
½ [m * v^2] = u * (m*g) *S  { F = m*g }
½ v^2 = u * g * S
S = v^2 / (2 * u * g)
Substituting the values:
v = 15 m/s
u = 0.2
g = 9.8 m/s^2
S = [(15)^2] / [2 * (0.2) * (9.8)]
S = 225 / 3.92
S = 57.39 m
S ~ 57.4 m
So the car will travel 57.4 meter? before coming to rest
```
2 years ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions