A 12 kg battery on a frictionless bench is accelerated by a light string that passes over a light pulley. The string is attached to a 6.0 kg paint can hanging over the table. When the paint can be released
a) what is the magnitude of the acceleration of the battery?
b) what is the magnitude of the net force on the battery?
c) what is the magnitude of the net force on the paint can?
a) what is the magnitude of the acceleration of the battery?
b) what is the magnitude of the net force on the battery?
c) what is the magnitude of the net force on the paint can?
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a) Force accelerating the system = weight of the paint can: Wp = (Mp)g
Ignoring the mass of the string and pulley:
Wp = (Mp + Mb)a
a = Wp/(Mp + Mb) = (Mp)g/(Mp+Mb)
a = 6g/(6+12) = g/3
If g is taken to be 9.81 m/s², acceleration of all parts of the system, not just the battery, is:
a = 9.81/3 = 3.27 m/s²
b) Net force on the battery is the tension on the string which is the accelerating force:
T = F = ma = 12*3.27 = 39.24 N
c) Net force on the paint can is the accelerating force on the can,
F = ma = 6*3.27 = 19.62 N
which is also equal to the weight of the can minus the tension in the string:
F = W-T = 6*9.81 - 39.24 = 19.62N
Ignoring the mass of the string and pulley:
Wp = (Mp + Mb)a
a = Wp/(Mp + Mb) = (Mp)g/(Mp+Mb)
a = 6g/(6+12) = g/3
If g is taken to be 9.81 m/s², acceleration of all parts of the system, not just the battery, is:
a = 9.81/3 = 3.27 m/s²
b) Net force on the battery is the tension on the string which is the accelerating force:
T = F = ma = 12*3.27 = 39.24 N
c) Net force on the paint can is the accelerating force on the can,
F = ma = 6*3.27 = 19.62 N
which is also equal to the weight of the can minus the tension in the string:
F = W-T = 6*9.81 - 39.24 = 19.62N