Question

A car of 900 kg mass is moving at the velocity of 60 km/hr. It is brought into rest at 50 meter distance by applying a brake. Now, calculate the force required to stop the car.

Answer 1

Answer: $-2502N$

Explanation:

$(V_2)^2=(V_1)^2+2ad$

where;

$V_2$ = final velocity = 0

$V_1$ = initial velocity = 60 km/h = 16.67 m/s

$a$ = acceleration

$d$ = distance

First all of, because acceleration is given in m/s and not km/h, you need to convert 60km/h to m/s. Our conversion factors here are 1km = 1000m and 1h = 3600s

$60km/h(\frac{1000m}{1km} )(\frac{1h}{3600s} )=16.67m/s$

Solve for a;

$(V_2)^2=(V_1)^2+2ad$

Begin by subtracting $(V_1)^2$

$(V_2)^2-(V_1)^2=2ad$

Divide by 2d

$\frac{(V_2)^2-(V_1)^2}{2d} =a$

Now plug in your values:

$a=\frac{(0)^2-(16.67 m/s)^2}{2(50m)}$

$a=\frac{0-277.89m^2/s^2}{100m}$

$a=-2.78m/s$

If you're wondering why I calculated acceleration first is because in order to find force, we need 2 things: mass and acceleration.

$F=ma$

m = mass = 900kg

a = acceleration = -2.78m/s

$F=(900kg)(-2.78m/s)\\F=-2502N$

It's negative because the force has to be applied in the opposite direction that the car is moving.