Question

How large an average force is required to stop a 1400-kg car in 5.0 s if the car's initial speed is 25 m/s?

a)2000 N

b)3500 N

c)9000 N

d)7000 N

Answer 1

Answer on Question 59420, Physics, Mechanics, Relativity

Question:

How large an average force is required to stop a $1400\,\mathrm{kg}$ car in $5.0\,\mathrm{s}$ if the car's initial speed is $25\,\mathrm{ms}^{-1}$?

a) $2000\,\mathrm{N}$

b) $3500\,\mathrm{N}$

c) $9000\,\mathrm{N}$

d) $7000\,\mathrm{N}$

Solution:

We can find an average force that required to stop the car from the definition of the impulse:

$$\Delta p = F_{avg} \Delta t,$$ $$m \Delta v = F_{avg} \Delta t,$$ $$m(v_{final} - v_{initial}) = F_{avg} \Delta t$$

here, $m$ is the mass of the car, $v_{initial}$ is the initial speed of the car, $v_{final}$ is the final speed of the car, $F_{avg}$ is the average force that required to stop the car, $\Delta t$ is the change in time.

Then, from the last formula we can calculate the average force:

$$F_{avg} = \frac{m(v_{final} - v_{initial})}{\Delta t} = \frac{1400\,\mathrm{kg} \cdot (0\,\mathrm{ms}^{-1} - 25\,\mathrm{ms}^{-1})}{5.0\,\mathrm{s}} = -7000\,\mathrm{N}.$$

The sign minus indicates that the average force directed opposite to the direction of the motion of the car. The magnitude of the average force is equal to $7000\,\mathrm{N}$.

Answer:

d) $7000\,\mathrm{N}$

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