A person slides a box down a ramp. The box starts from rest 2m above the lowest point. The ramp is frictionless. Once the box reaches the bottom there is friction. For every 1m the box travels on the flat part increases by 100 J. Where does the box stop? The box has mass of 5 kg.

This question involves the concepts of the law of conservation of energy, work done, Newton's Second law of motion, and equation of motion.

The box stops at a distance of "1.02 m".

First, we will use the law of conservation of energy to find out the final velocity when the box reaches the flat part.

[tex]Loss\ of\ Potential\ Energy=Gain\ in\ Potential\ Energy\\mgh=\frac{1}{2}mv^2\\\\2gh=v\\v=\sqrt{2gh}[/tex]

where,

v = speed = ?

g = 9.81 m/s²

h = height = 2 m

Therefore,

[tex]v=\sqrt{(2)(9.81\ m/s^2)(2\ m)}[/tex]

v = 6.26 m/s

Now, we will use the formula of work done to find out the force applied to the box to stop the motion:

[tex]W = Fd[/tex]

where,

W = Work Done = 100 J

F = Force applied = ?

d = displacement = 1 m

Therefore,

[tex]100\ J = F(1\ m)\\F = 100\ N\\[/tex]

Now, we will use Newton's Second Law of Motion to find out the deceleration of the person:

[tex]F=ma[/tex]

where,

m = mass = 5 kg

a = decceleration =?

Therefore,

[tex]100\ N = (5\ kg)a\\\\a=\frac{100\ N}{5\ kg}\\\\a = - 20\ m/s^2[/tex]

negative sign shows deceleration.

Now, we will use the third equation of motion to find the distance at which the box stops.

[tex]2as=v_f^2-v_i^2\\2(-20\ m/s^2)s=(0\ m/s)^2-(6.26\ m/s)^2\\\\s=\frac{40\ m/s^2}{39.24\ m^2/s^2}[/tex]

s = 1.02 m

Learn more about equations of motion here:

brainly.com/question/20594939?referrer=searchResults

The attached picture shows the equations of motion in the horizontal and vertical directions.