A box is moving at 4.00 m/s along a horizontal hard packed snow surface. The coefficient of friction is 0.05. How far does the box go before stopping?

Question

A box is moving at 4.00 m/s along a horizontal hard packed snow surface. The coefficient of friction is 0.05. How far does the box go before stopping?

Thanks to anyone who responds and please show work!

Answer 1

What we have here is a nice little motion problem, an object goes from some initial velocity to a dead stop, there's gravity and friction, etc. Whenever you see a problem like this, pay attention to whether or not TIME is involved anywhere (Does the problem give you a time or ask for a time?). IF TIME IS INVOLVED: you're going to have to solve the problem using kinematics, or possibly momentum and impulse. IF TIME IS NOT INVOLVED: you can most likely just solve it using conservation of energy.

So here, we DO NOT have any time involved in the problem. We can use conservation of energy:

The problem starts with an object moving at 4 m/s, so it has KINETIC ENERGY

E_initial = KE = ¹/₂· m·v²

Where v=4.0 m/s.

But then, the problem ends with the box coming to a stop. So where did that energy go? It must have gone somewhere.

The energy was taken from the box by the friction applied to the box. What has happened is the FRICTION did WORK on the box, robbing it of its kinetic energy. If you are curious, the energy was released in the form of heat and sound. (you can hear the box sliding on the snow).

The work done by the friction must be equal to the energy the box originally had:

Work = Force·distance = ƒ·D

and

ƒ = µ·m·g

With ƒ being the frictional force, µ being the coefficient of kinetic friction, g being 9.81 m/s², m being the mass of the box, and D being the distance slid.

If the work done is equal to the initial energy "E_initial", we can set the expressions equal:

E_initial=Work

¹/₂· m·v² = ƒ·D

¹/₂· m·v² = µ·m·g·D Starting here: divide both sides by m·

¹/₂··v² = µ·g·D Divide by µ and g

¹/₂·( v²) / (µ·g) =·D

So that's velocity squared, divided by the product of µ and g, THEN multiplied by 0.5

So, the answer I got is:

D = 16.31 meters

Answer 2

We can use therem abut changes of kinetic energy: 0 - mv₀²/2 = - µmgs, where µ is coefficient of friction, d is distance.

So s = v₀²/(2µg) = 4²/(2·0.05·9.81) = 16.31 m