F = 125 N
mass = (125+305)/10 = 43 kg
a = F/m = 125/43 = 2.9 m/s^2
The weight of the block at rest on a frictionless table is 305 N. This block is attched by a string to another mass that hangs off a frictionless pulley. And the weight of the hanging block is 125 N. Ignoring all frictional effects and assuming the pulley to be massless, find the acceleration of the two blocks. Use g = 10 m/s^2 in all calculations
I do not know what formula to use please help.
2 answers
Sum of X forces = T = m1a = (30.5kg*a)
Sum of Y forces = 125N - T = m2a = (12.5kg*a)
125N - (30.5kg*a) = (12.5kg*a)
125N = 43kg*2a
2.85m/s^2 = 2a
1.425m/s^2 = a
Sum of Y forces = 125N - T = m2a = (12.5kg*a)
125N - (30.5kg*a) = (12.5kg*a)
125N = 43kg*2a
2.85m/s^2 = 2a
1.425m/s^2 = a
0 answers