Question

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

Answer 1

Answer: I think instead of 4 sec there will be 5 sec... Plz check

A mass of 3 kg descending vertically downwards supports a mass of 2 kg by means of a light string passing over a pulley. At the end of 5 seconds, the string breaks. How high now will that 2 kg mass go?

Explanation:

To solve this question we shall first find the upwards acceleration of the 2 kg block. The acceleration would be

F/m = (30–20)/2 =10/2 = 5 m/s²

At the end of 5 seconds the velocity of the block will be

v = u+ at → v = at (u=0) → v = 5*5 → v = 25m/s

Now, the string breaks and the block keeps moving up decelerating at 10 m/s². So, first we will find the time it keeps moving up.

v = u + at → 0= 25 — (10*t) → t= 2.5 seconds

Now putting the values in s= ut + (1/2)at² we shall find the displacement of block.

Answer 2

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