Question

A block of mass 5kg is being pulled up by a constant
force of 90 N starting at ground from rest. 2 sec later, a
part of the block having mass 3kg falls off. Find the
distance between the two parts when 3kg part hits the
ground,​

Answer 1

Answer:

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Answer 2

Given:

Mass of block = 5 kg

Force = 90 N

Mass of the block that falls off = 3 kg

To find:

Distance between the two parts when the 3 kg part hits the ground

Solution:

As we know that:

S = ut + 1/2 at²

Here F = 90N = m *a

a = 90/ 5 = 18 m/s²

Then S = 0 + 1/2 * 18 * 2²

As the block was at rest initially.

S = 36 m

After t=2 sec, a 3 kg block falls off.

so, mass of the remaining block = 5 kg

3 kg will take how much time to reach the ground:

S = ut + 1/2 at²

36 = 0 + 1/2 * 9.8 * t²

t = √7.35

In time t, the 2 kg block must have covered a distance S with a 90N force acting on it:

S = ut + 1/2 at²

And F = m*a

a = 90/2 = 45m/s²

v = u + at

v = 0 + 18 * 2

v = 36 m/s

S = 36 * 2.7 + 1/2*45*7.35

S = 97.2+165.375

S = 262.575 m

Total distance between the two blocks = 262.575 + 36 = 298.575 m

Therefore the distance between the two parts is 298.575 m.