What is the magnitude of the upward acceleration of the load of bricks?

Question

A load of bricks with mass = 15.0 kg hangs from one end of a rope that passes over a small, frictionless pulley. A counterweight of mass m2= 28.0kg is suspended from the other end of the rope, as shown in the figure. The system is released from rest.What is the magnitude of the upward acceleration of the load of bricks?
Take the free fall acceleration to be = 9.80m/s^2 .What is the tension in the rope while the load is moving?
Take the free fall acceleration to be = 9.80m/s^2 .

Answer 1

Look at a FBD of the bricks and of the counterweight. Since they are connected over a frictionless pulley the accelerations and tension in the cables will be equal in magnitude

bricks
15*a=T-15*9.8

counterweight
28*a=28*9.8-T

solve for T first

T/15-9.8=a
9.8-T/28=a


T/15-9.8=9.8-T/28
T(1/15+1/28)=2*9.8
T=191 N

a=2.96 m/s^2

j

Answer 2

the load of the bricks is 15.4 x 9.8 = 151 N of the load is 29 x 9.8 = 284 N the internet is the adaptation of 133 N, that's the stress interior the rope. entire mass = 15.4+29 = 40 4.4 kg F = ma a = F/M = 133/40 4.4 = 3.00 m/s²

Answer 3

Find the FBD. From that we can conduct that the only direction were using is the j-hat (y) direction.
Start with Fy=m1ay and Fy=m2ay these can be conducted at the same time
Fy=m1ay (Plug in for Fy) (Tension pulls up and Gravity is negative making +T and -m1g)
T-mg=m1ay (Divide m1 on both sides) [odu83 wrote this very poorly]
(T/m)-g=ay
Fy=m2ay (Plug in for Fy) (Tension pulls down and Gravity is positive making +mg and -T)
m2g-T=m2ay (Divide m2 from both sides)
g-(T/m2)=ay
Now its time for algebra (its simple but annoying)
ay=ay which means
(T/m)-g=g-(T/m) (We must do this because there are two variables)
Plug in find T
(T/15)-9.8=9.8-(T/28)
T((1/15)+(1/28))=2(9.8)
.102T=19.6
T=(19.6/.102)
T=192.16
Plug in T to find a
(T/m1)-g=ay
((192.16)/(15))-9.8=ay
ay=3.01 m/s^2

Answer 4

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Answer 5

how did you solve for a