Find the acceleration of the 500 g block in the figure (5−E18).
Answers
Given:
Refer the attachment diagram for data
m1 = 100 g = 0.1 kg
m2 = 500 g = 0.5 kg
m3 = 50 g = 0.05 kg
The free-body diagram for the system is given in attachment.
From the free-body diagram of the 500 g block,
T + 0.5a − 0.5g = 0 …..(i)
From the free-body diagram of the 50 g block,
T1 + 0.05g − 0.05a = a ….(ii)
From the free-body diagram of the 100 g block,
T1 + 0.1a − T + 0.5g = 0 ….(iii)
From equation (ii),
T1 = 0.05g + 0.05a …..(iv)
From equation (i),
T1 = 0.5g − 0.5a …..(v)
Equation (iii) becomes
T1 + 0.1a − T + 0.05g = 0
From equations (iv) and (v), we get:
0.05g + 0.05a + 0.1a − 0.5g + 0.5a + 0.05g = 0
0.65a = 0.4 g
⇒a=[0.4/0.65]g
=[40/65]g=[8/13]g downward
So, the acceleration of the 500 gm block is [8g/13] (downward).
Answer:8g/13
Explanation:If you go through that question you may find some information that is given.
Let the acceleration be a .
Now,
m1=500g
m2 =100g
m3=50g
From the data of 500g block-
500.g - T1=500a.......(i)
From the data of 50g block-
T2-50.g = 50a
And in case of 100 gram block ,
We know
Fnet= ma .
ATQ,
T1-T2-W=ma
>500(g-a)-50(a+g)-100.g.sin30°=100a
>400g=650a
>a=8g/13 ✓✓