Distance a Block Travels Across a Frictional Surface?

Q: A block with a mass of 0.50kg is released from the top of a frictionless slide with a height of 0.50m. At the bottom of the slide the block travels some distance, Δx, across a horizontal surface with a friction coefficient of μk=0.25. Find Δx.

A: At the bottom of the slide,

v² = 2gh = 2 * 9.8m/s² * 0.50m = 9.8 m²/s²

so its KE = ½mv² = ½ * 0.50kg * 9.8m²/s² = 2.45 J

This energy is dissipated by friction work:

work = friction force * distance = µmg * Δx

2.45 J = 0.25 * 0.50kg * 9.8m/s² * Δx

Δx = 2.0 m

Hope this helps!

At the top of the slide the block will have a gravitational potential energy = m.g.h

where m is the mass; g is the acceleration due to gravity {=9.8 m/s²} and h is the vertical height

At the bottom of the slide, all that g.p.e. will be converted into kinetic energy. The slide is frictionless, so none of the original energy will be lost to friction on the way down.

k.e. of block at bottom of slide = (0.50 kg * 9.8 m/s² * 0.50 m) = 2.45 joules

Now we can use the work~energy theorem.

Work done by friction to bring the block to rest = change in k.e. = 2.45 J

OK, now we need to involve the relationship between the coefficient of friction (μ), the friction force (Ff) and the normal force (Fn) {The normal force is the force at right angles to the friction surface.}

Ff = μ * Fn

On a horizontal surface, Fn is just equal in magnitude to the weight force (mg)

Friction force on horizontal surface = μk * mg = (0.25 * 0.50 * 9.8) = 1.225 newtons

Work done = force * distance

2.45 J = 1.225 N * Δx

Δx = (2.45 / 1.225) = 2 m