University Physics I - Classical Mechanics, 2019

154 CHAPTER 7. IMPULSE, WORK AND POWER
7.7.2 Work, energy and the choice of system: dissipative case
Consider again the situation shown in Figure 7.5. Letm 1 =1kg,m 2 =2kg,andμ k =0.3. Use the
solutions provided in Section 6.3 to calculate the work done by all the forces, and the changes in all
energies, when the system undergoes a displacement of 0.5 m, and represent the changes graphically
using bar diagrams like the ones in Figure 7.3 (for system A and B separately)
Solution
From Eq. (6.32), we have
a = m 2 − μ k m 1
m 1 + m 2
g =5.55 m s 2
F t = m 1m 2 (1 + μ k )
g =8.49 N (7.34)
m 1 + m 2
We can use the acceleration to calculate the change in kinetic energy, since we have Eq. (2.10) for
motion with constant acceleration:
(
vf 2 − v2 i =2aΔx =2× 5.55 m )
s 2 × 0.5m =5.55 m2
s 2 (7.35)
so the change in kinetic energy of the two blocks is
ΔK 1 = 1 2 m 1
(
v
2
f − v 2 i
)
=2.78 J
ΔK 2 = 1 2 m 2
(
v
2
f − v 2 i
)
=5.55 J (7.36)
We can also use the tension to calculate the work done by the external force on each system:
W ext,A = Fr,1Δx t =(8.49 N) × (0.5m)=4.25 J
W ext,B = Fr,2Δy t =(8.49 N) × (−0.5m)=−4.25 J (7.37)
Lastly, we need the change in the gravitational potential energy of system B:
ΔUB G = m 2 gΔy =(2kg)×
(9.8 m )
s 2 × (−0.5m)=−9.8 J (7.38)
and the increase in dissipated energy in system A, which we can get from Eq. (7.28):
ΔE diss = −Fs,1Δx k = μ k Fs,1Δx n = μ k m 1 gΔx =0.3 × (1 kg) ×
(9.8 m )
s 2 × (0.5m)=1.47 J (7.39)
We can now put all this together to show that Eq. (7.20) indeed holds:
W ext,A =ΔE A =ΔK 1 +ΔE diss =2.78 J + 1.47 J = 4.25 J
W ext,B =ΔE B =ΔK 2 +ΔUB G =5.55 J − 9.8J =−4.25 J (7.40)