283Two questions
Akash Anand For second one ..refer to Laws of Motion_Day-3 pptUpvote·0· Reply ·2013-07-01 23:15:38- Akash Anand For first one..draw tension table and it will be easy
5 Answers
283
9961) acceleration = 0.4m/s2
Tension in cord connecting m2=0.32 N
Tension in cord connecting m1=0.16 N
- Akash Anand check it again ..i guess u had made some mistakes.
996(2) For the mass m1 to be at rest, T1 = m1g …(1)
For the sum of forces across the pulley 2 carrying masses m2 and m3 to be zero, we have
2T2 = T1 …(2)
Let the acceleration of mass m2 be ‘a’. Applying Newton’s second law, we have
T2 – 20 = 2a …(3)
Applying Newton’s second law on mass m3, we have
30 - T2 = 3a …(4)
Solving equations (3) and (4), we get
a = 2 ms-2
T2 = 24 N
Putting in (2), we get
T1 = 48 N
Putting in (1), we get
m1 = 4.89 kg
9961) F-2T = m2a ........(1)
T = m1a.............(2)
on solving we get
a=0.28 m/s2
Tension in cord connecting m2=0.344 N
Tension in cord connecting m1=0.172 N
- Akash Anand Acceleration for both the blocks wont be same.
- Swarna Kamal Dhyawala ohh yeah
T = 2m1a
- Swarna Kamal Dhyawala T= m1a .......(2)
51Tension in the cord connecting m2 and pulley = 12/35 N
Acceleration of m2= 2/7 ms-2