Set 3 Problem number 14

• Problem

• Solution

• Generalized Solution

• Explanation in terms of Figure(s), Extension

• Figure(s)

Problem

• If the coefficient of friction between the mass and the horizontal surface over which it travels is .03, then how far will the object travel before coming to rest?

Solution

By constructing a force vs. stretch graph we can find the total work required to stretch the band as the area under the associated curve, between stretch=0 and stretch= 5.5 cm, to be about

• work to stretch = area under curve = 1.1885 Joules (approx)

( reasonable trapezoidal approximation works well here). 

• (University Physics students note:  A good model for the force function if F(x) = 300 * x^.72, where x is stretch in meters when y is in Newtons.  This function can be integrated from x=0 to x = 5.5 cm to obtain the work (be sure to change 5.5 cm to meters).

Neglecting thermal losses (which tend to be significant for a rubber band) we see that the PE of the system consisting of the stretched rubber band and the mass is

• PE = 1.1885 Joules.

If all of this PE goes into the KE of the mass, then we will have .5 m v^2 = 1.1885 Joules, with m = .114 kg. 

• Solving .5 m v^2 = KE for v we get

• v = `sqrt(2 * KE / m); substituting we see that
• velocity attained = v = `sqrt(2 * 1.1885 Joules / ( .114 kg) ) = 3.22 m/s.

If there is a frictional force the KE will be dissipated as the object does work against this force. 

• In this case the original KE will be equal to the work `dW = fFrict `ds, where fFrict is the frictional force and `ds the distance. 
• If we know fFrict and `dW, we can solve to obtain `ds = `dW/fFrict.

Since the object is on a horizontal surface under the influence only of friction, gravity and the normal force, the only vertical forces are the object's weight and the normal force, which must be equal and opposite. 

• The normal force and weight therefore both have magnitude .114 kg * 9.8 m/s^2 = 1.1172 Newtons. 
• The frictional force therefore has magnitude fFrict = coeff. of friction * normal force = .03 * 1.1172 Newtons = .033 Newtons.

The distance will therefore be `ds = `dW / fFrict = 1.1885 Joules / .033 Newtons = 36.01 meters.

If the object moves up an incline it will continue until all the original elastic PE in the rubber band has been changed to gravitational PE.  

• The rubber band imparts a KE (and also increases the object's PE before the rubber band goes slack); this KE  then continues to decrease as the gravitational PE increases. 

• If the displacement of the mass in the vertical direction is `dy, we see that `dy and the force exerted by gravity are in parallel but opposite directions, so that gravity does negative work on the object; the object therefore does positive work against gravity, with the expected resulting decrease in KE.

• The object stops when the change in the gravitational PE is equal to the original elastic PE in the rubber band. 
• The change in the gravitational PE is the work done against gravity by the object as its vertical position changes by `dy. 
• This work is equal to the weight of the object multiplied by `dy: 

• `dPE = weight * `dy = 1.1172 Newtons * `dy. 

• If `dPE = 1.1885 Joules, then

• `dy = 1.1885 J / ( 1.1172 N) = 1.06382 meters.

If we consider the gravitational force component Fparallel = 1.1172 Newtons * sin( 8.2 deg) = .159 Newtons down the incline, then if `dsParallel is the displacement of the object along the incline, the work done against gravity is

• work by gravity = .159 Newtons * `dsParallel = 1.1885 Joules. 
• We therefore see that the displacement along the incline is

• displacement along incline = `dsParallel = 1.1885 Joules / ( .159 Newtons) = 7.474 meters.

• Note that the vertical component of the corresponding displacement vector is 7.474 meters * sin( 8.2 deg) = 1.06382 meters, agreeing with the vertical displacement already calculated.