8. A mass of 5 kg descending vertically, draws up amass of 3 kg by mea

1.	8. A mass of 5 kg descending vertically, draws up amass of 3 kg by means of a light string passing overa pulley. At the end of 4 s, the string breaks. Howmuch higher the 3 kg mass would go ?
Answer» Given :- Mass of big block = M = 5 KG Mass of SMALL block = m = 3 Kg Force on big block = F = 5g Force on small block = f = 3g Time = t = 4 s ∆F = (M+m)a F - f = (5+3)a a = 2g/8 a = g/4 Using FIRST equation of Motion, v = at (Since Initial Velocity is Zero.) v = g/4 × 4 v = g As we know that, h = v²/2g h = g²/2g h = g/2 Taking 'g' = 9.8 ms-² h = 4.9 m Hence, Mass 3 kg will go = h = 4.9 m

8. A mass of 5 kg descending vertically, draws up amass of 3 kg by means of a light string passing overa pulley. At the end of 4 s, the string breaks. Howmuch higher the 3 kg mass would go ?

Answer»

Given :-

Mass of big block = M = 5 KG

Mass of SMALL block = m = 3 Kg

Force on big block = F = 5g

Force on small block = f = 3g

Time = t = 4 s

∆F = (M+m)a

F - f = (5+3)a

a = 2g/8

a = g/4

Using FIRST equation of Motion,

v = at (Since Initial Velocity is Zero.)

v = g/4 × 4

v = g

As we know that,

h = v²/2g

h = g²/2g

h = g/2

Taking 'g' = 9.8 ms-²

h = 4.9 m

Hence,

Mass 3 kg will go = h = 4.9 m

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8. A mass of 5 kg descending vertically, draws up amass of 3 kg by means of a light string passing overa pulley. At the end of 4 s, the string breaks. Howmuch higher the 3 kg mass would go ? ​

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8. A mass of 5 kg descending vertically, draws up amass of 3 kg by means of a light string passing overa pulley. At the end of 4 s, the string breaks. Howmuch higher the 3 kg mass would go ?