Problem#1
A bead at the bottom of a bowl is one example of an object in a stable equilibrium position. When a physical system is displaced by an amount x from stable equilibrium, a restoring force acts on it, tending to return the system to its equilibrium configuration. The magnitude of the restoring force can be a complicated function of x. For example, when an ion in a crystal is displaced from its lattice site, the restoring force may not be a simple function of x. In such cases we can generally imagine the function F(x) to be expressed as a power series in x, as F(x) = –(k1x + k2x2 + k3x3 + . . .). The first term here is just Hooke’s law, which describes the force exerted by a simple spring for small displacements. For small excursions from equilibrium we generally neglect the higher order terms, but in some cases it may be desirable to keep the second term as well. If we model the restoring force as F = –(k1x + k2x2), how much work is done in displacing the system from x = 0 to x = xmax by an applied force –F?Answer:
The work done by the applied force is
W = ∫if Fapplieddx
W = –∫0xmax[–(k1x + k2x2)]dx
W = ∫0xmaxk1xdx + ∫0xmax k2x2)dx
W = ½ k1x2max + k2x3max/3
Problem#2
A traveler at an airport takes an escalator up one floor, as in Figure 1. The moving staircase would itself carry him upward with vertical velocity component v between entry and exit points separated by height h. However, while the escalator is moving, the hurried traveler climbsthe steps of the escalator at a rate of n steps/s. Assume that the height of each step is hs. (a) Determine the amount of chemical energy converted into mechanical energy by the traveler’s leg muscles during his escalator ride, given that his mass is m. (b) Determine the work the escalator motor does on this person.
(a) The work done by the traveler is
W = mghsN
where N is the number of steps he climbs during the ride.
with N = (time on escalator)(n)
where, time on escalator = h/(vertical velocity of person)
and vertical velocity of person = v + nhs
then, N = hn/(v + nhs)
and the work done by the person becomes
W = mghshn/(v + nhs)
(b) The work done by the escalator is
We = Pt = Fvt = mgvt
where t = h/(v + nhs) as above.
Thus, We = mgvh/(v + nhs)
As a check, the total work done on the person’s body must add up to mgh, the work an elevator would do in lifting him.
It does add up as follows:
∑W = Wperson + We
= mghshn/(v + nhs) + mgvh/(v + nhs)
∑W = mgh(v + nhs)/(v + nhs) = mgh
Problem#3
A mechanic pushes a car of mass m, doing work W in making it accelerate from rest. Neglecting friction between car and road, (a) what is the final speed of the car? During this time, the car moves a distance d. (b) What constant horizontal force did the mechanic exert on the car?
Answer:
(a) the final speed of the car. We use
∆K = ∑W
Kf – Ki = ∑W
½ mv2 – 0 = W
v2 = 2W/m
v = √(2W/m)
(b) constant horizontal force did the mechanic exert on the car is
W = F.v = Fxd
Fx = W/d
Problem#4
A 5.00-kg steel ball is dropped onto a copper plate from a height of 10.0 m. If the ball leaves a dent 3.20 mm deep, what is the average force exerted by the plate on the ball during the impact?
Answer:
During its whole motion from y = 10.0 m to y = -3.20 mm, the force of gravity and the force of the plate do work on the ball. It starts and ends at rest
∑W = ∆K
∑W = Kf – Ki
mg∆y cos00 + Fp∆x cos1800 = 0
Fp = mg∆y/∆x
Fp = (5.00 kg)(9.80 m/s2)(10 m)/(3.20 x 10-3 m)
Fp = 1.53 x 105 N upward
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