Question

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?

Answer 1

Answer:

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

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 $v^2-u^2=2as$

$0^2-12.4^2=2\times a\times 78$

$a=-0.9856m/sec^2$

Magnitude of acceleration $a=0.9856m/sec^2$

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

Answer 2

Answer:

1419.01436 N

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-12.4^2}{2\times 78}\\\Rightarrow a=-0.98564\ m/s^2$

The force on the car

$F=ma\\\Rightarrow F=1420\times -0.98564\\\Rightarrow F=-1419.01436\ N$

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