Question

at a certain moment, the block are set in motion from the position shown. The maximum height that block of mass 3M will go up to is
1) 2h/3 2) 5h/4
3) 7h/4 4) 2h​

Answer 1

1. mark the tensions in the string...
2. whole string will have same tension as the same massless string is present
3. use Fnet=Fdriving-Fopposing=ma formula
4. the whole system is moving together..so acceleration is same for both blocks

let's solve!!

For 5kg block..

• 5g-T=5a-----(1)

For 3kg block...

• T-3g=3a------(2)

☆adding equation 1 and 2

• 5g-3g=8a
• 2g=8a (g=10m/s2)
• a=20/8
• a=5/2
• a=2.5m/s^2

see the attachment!!with solving:-

For 5kg ...

• ${v}^{2} = {u}^{2} + 2as$
• $\frac{1}{2} m {v}^{2} = mg h_{1} \\ {v }^{2} =2 gh_{1}$

For 3kg..

• ${v}^{2} = {u}^{2} + 2a(h - h_{1} )$
• $0 = 2g h_{1} + 2g(h - h_{1})$
• $- 2g h_{1} = 2g(h - h_{1})$

the maximum height it goes is 2h