Principle of Work and Energy for Particles in a Plane (example continued)
Strategy: Apply the principle of work and energy to determine the speed of the blocks
after B has dropped 2 ft. Define the system to be block A + the pulley + block B. The
work done on the system W1-2, from position 1 to position 2 equals the change in kinetic
energy of the system T2 – T1 such as:
W1-2 = T2 - T1
Show the forces that do work during displacement, d, of the system from position (1) to
position (2). See the figure below.
The work done on the system, W1-2 = WB d – μ WA d
The change in kinetic energy T1-2 = T2 – T1 = ½ ma v22 + ½ mbv22 - 0
WB d – μ WA d = ½ ma v22 + ½ mb v22
v22 = 2 ( WB – μ WA ) d / (ma + mb ) = 2g (WB – μ WA )(2) / (WA + WB)
v22 = √ [2g (WB – μ WA )(2) / (WA + WB)] = √ [ 64.4 (16.1 – 6.44)(2)/(48.3) ]
v2 = 5.075 ft/sec (result) v1 = v2 since cord is inextensional.
Click here for another example.
Return to Notes on Dynamics
Copyright © 2019 Richard C. Coddington All rights reserved.