Question

An 800 kg car moving at 30 m/s brakes and skids to a stop over surface with coefficient of friction 0.60 make diagram and find force of friction and distance to stop

Answer 1

Answer on Question #45829, Physics, Mechanics | Kinematics | Dynamics

Question:

An $800\mathrm{kg}$ car moving at $30~\mathrm{m / s}$ brakes and skids to a stop over surface with coefficient of friction 0.60 make diagram and find force of friction and distance to stop

Answer:

Newton's second law of motion:

$$m a = F _ {f r}$$

Newton's first law of motion:

$$N = m g$$

Friction force equals

$$F _ {f r} = \mu N = \mu m g = 0. 6 \cdot 9. 8 \cdot 8 0 0 = 4 7 0 0 N$$

where $\mu$ - coefficient of friction.

Therefore:

$$a = \frac {\mu m g}{m} = \mu g$$

distance to stop equals:

$$d = \frac {v ^ {2}}{2 a} = \frac {v ^ {2}}{2 \mu g} = \frac {3 0 ^ {2}}{2 \cdot 0 . 8 \cdot 9 . 8} = 7 7 m$$

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