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A 1420-kg car is traveling with a speed of 12.4 m/s. What is the magnitude of the horizontal net force that is required to bring the car to a halt in a distance of 78.0 m?

2 Answers

5 votes

Answer:

So net force will be F = 1420×0.985 = 1399.61 N

Step-by-step explanation:

We have given mass of the car m = 1420 kg

Initial velocity of the car u = 12.4 m/sec

As the car finally stops final velocity v = 0 m/sec

Distance after which car is halt s = 78 m

Now according to third equation of motion
v^2-u^2=2as


0^2-12.4^2=2* a* 78


a=-0.9856m/sec^2

Magnitude of acceleration
a=0.9856m/sec^2

So net force will be F = 1420×0.985 = 1399.61 N

User Fabian Sierra
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5.6k points
2 votes

Answer:

1419.01436 N

Step-by-step explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration


v^2-u^2=2as\\\Rightarrow a=(v^2-u^2)/(2s)\\\Rightarrow a=(0^2-12.4^2)/(2* 78)\\\Rightarrow a=-0.98564\ m/s^2

The force on the car


F=ma\\\Rightarrow F=1420* -0.98564\\\Rightarrow F=-1419.01436\ N

Magnitude of the horizontal net force that is required to bring the car to a halt is 1419.01436 N

User Kcsurapaneni
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5.2k points