Problem#1
At time ti, the kinetic energy of a particle is 30.0 J and the potential energy of the system to which it belongs is 10.0 J. At some later time tf, the kinetic energy of the particle is 18.0 J. (a) If only conservative forces act on the particle, what are the potential energy and the total energy at time
tf ? (b) If the potential energy of the system at time tf is 5.00 J, are there any nonconservative forces acting on the particle? Explain.
Answer:
(a) the potential energy at time tf, we use
Ki + Ugi = Kf + Ugf
30.0 J + 10.0 J = 18.0 J + Ugf
Ugf = 22.0 J
and the total energy at time tf is
Ef = Kf + Ugf = 18.0 J + 22.0 J = 40.0 J
(b) Yes, because ∆Emech = ∆K + ∆U is not equal to zero. For corservative force ∆K + ∆U = 0.
Problem#2
In her hand a softball pitcher swings a ball of mass 0.250 kg around a vertical circular path of radius 60.0 cm before releasing it from her hand. The pitcher maintains a component of force on the ball of constant magnitude 30.0 N in the direction of motion around the complete path. The speed of the ball at the top of the circle is 15.0 m/s. If she releases the ball at the bottom of the circle, what is its speed upon release?
Answer:
The distance traveled by the ball from the top of the arc to the bottom is πR . The work done by the
non-conservative force, the force exerted by the pitcher, is
∆E = F∆rcos00 = F(πR)
We shall assign the gravitational energy of the ball-Earth system to be zero with the ball at the bottom of the arc
Then,
∆Emech = ∆K + ∆U = ½ mvf2 – ½ mvi2 + mgyf – mgyi
F(πR) = ½ mvf2 – ½ mvi2 + mgyf – mgyi
½ mvf2 = F(πR) + ½ mvi2 + mgyi
½ (0.250 kg)vf2 = (30.0 N)π(0.600 m) + ½ (0.250 kg)(15.0 m/s)2 + (0.250 kg)(9.80 m/s2)(1.20 m)
0.250kgv2 = 175.17 kgm2/s2
v = 26.5 m/s
Problem#3
An electric scooter has a battery capable of supplying 120 Wh of energy. If friction forces and other losses account for 60.0% of the energy usage, what altitude change can a rider achieve when driving in hilly terrain, if the rider and scooter have a combined weight of 890 N?
Answer:
Efficiency = useful output energy/total input energy
If friction forces and other losses account for 60.0% of the energy usage, then efficiency 40%,
40% = mg∆y/120Wh
∆y = (120 W)(3600 s)(0.400)/(890 N) = 194 m
Problem#4
The world’s biggest locomotive is the MK5000C, a behemoth of mass 160 metric tons driven by the most powerful engine ever used for rail transportation, a Caterpillar diesel capable of 5 000 hp. Such a huge machine can provide a gain in efficiency, but its large mass presents challenges as well. The engineer finds that the locomotive handles differently from conventional units, notably in braking and climbing hills. Consider the locomotive pulling no train, but traveling at 27.0 m/s on a level track
while operating with output power 1 000 hp. It comes to a 5.00% grade (a slope that rises 5.00 m for every 100 m along the track). If the throttle is not advanced, so that the power level is held steady, to what value will the speed drop? Assume that friction forces do not depend on the speed.
Answer:
As the locomotive moves on level track,
P = fvi
1000hp(746 W/1 hp) = f(27.0 m/s)
f = 2.76 x 104 N
As the locomotive moves up the hill at constant speed, its output power goes into internal energy
plus gravitational energy of the locomotive-Earth system:
Pt = mgy + f∆r = mg∆rsinθ + f∆r
P = mgvfsinθ + fvf
Then also
746000 W = (160000 kg)(9.80 m/s2)vf(5.00 m/100 m) + (2.76 x 104 N)vf
746000 W = (1.06 x 105 N)vf
vf = 7.04 m/s
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