Question

An advertisement claims that a car can stop on a dime. If the car is 850 kg and is traveling at 45.0 km/h, what is the magnitude of the net force needed to stop in the distance equal to the diameter of a dime (1.8 cm)

Answer 1

Answer:

Net force on the car will be $3689236.111N$      

Explanation:

We have given mass of the car m = 850 kg

Initial velocity of the car u = 45 km/hr $=45\times \frac{5}{18}=12.5m/sec$

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

Distance travel for stooping of car s = 1.8 cm = 0.018 m$v^2=u^2+2as$

From third equation of motion $v^2=u^2+2as$

So $0^2=12.5^2+2\times a\times 0.018$

$a=-4340.277m/sec^2$

So magnitude if acceleration will be $a=4340.277m/sec^2$

So net force on the car $F=ma=850\times 4340.277=3689236.111N$

So net force on the car will be $3689236.111N$