Two objects have the same kinetic energy and are moving in the same direction but, one of them has a much larg?

True, as the stopping distance is a function of Work done to stop the object/s.
Work done = F x d  = KE and since KE is the same for both objects
as is the F so "d" must also be the same. <= ANS

Let the force = -20 N
Acceleration = Force ÷ mass = (-20 ÷ m)

Final velocity^2 = Initial velocity^2 + (2 * a * d)
Final velocity^2 = 0 m/s
a = (-20 ÷ m)

0 = Initial velocity^2 + [2 * (-20 ÷ m) * d]
Initial velocity^2 = [2 * (20 ÷ m) * d]
Initial velocity^2 = (40 ÷ m) * d

Distance = Initial velocity^2 ÷ (40 ÷ m)


KE = ½ * m * v^2 = 100 J

KE2 has mass = 4 kg
½ * 4 * v^2 = 100
Divide both sides by ½ * 4
v^2 = 50
v = ?50 = ?(25 * 2)
v = 5 * ?2 m/s

Distance = Initial velocity^2 ÷ (40 ÷ m)
Distance = 50 ÷ (40 ÷ 4) = 5 m

KE1 has mass = 8 kg
½ * 8 * v^2 = 100
Divide both sides by ½ * 8
v^2 = 25
v = ?25
v = 5 m/s

Distance = Initial velocity^2 ÷ (40 ÷ m)
Distance = 25 ÷ (40 ÷ 8) = 5 m

If the mass is multiplied by 2, the velocity is divided by the ?2.
But the distance remains the same!!

Let the force to decrease the velocity from v to 0 m/s = 20 N
But the kinetic energy remains constant.

To stop either object, the work done by the force must decrease the kinetic energy from 100 Joules to 0 Joules!

The work causes the KE to decrease 100 Joules to 0 Joules, so the work = 100 Joules = 100 N * m

Work = Force * distance
The same force is applied to stop each object, so let the force = 20 N

100 N*m = 20 N * distance

100 N*m ÷ 20 N = distance
Distance = 5 m

The stopping distances for each object = 5 m

Now you know 2 methods to solve this type of problem.