Question

A couple of boxes with different masses, 4.00 kg and 1.40 kg , hang 0.400 m above the floor from the ends of a cord 6.20 mlong passing over a frictionless pulley. The boxes are released from rest with the heavy box falling and lifting up the light box.

Find the maximum height above the floor reached by the 1.40 kg  box.

Answer 1

Here, the acceleration of the system -

a = net force (N) / accelerated mass (kg)
a = [(4.00 - 1.40)kg x 9.80]N / [4.00 + 1.40]kg = 2.60 x 9.8 / 5.40 = 4.72 m/s²

So, upward velocity (v) acquired by 1.40 kg mass -  
v² = u² + 2ad .. (u = 0, d = 0.40m)
v² = 0 + (2 x 4.72 x 0.40m) = 3.78 (m/s)²

Now, 1.40 kg mass comes to rest after rising a further distance h, when all it's KE (½.mv²) is transferred to GPE (mgh).

½.mv² = mgh
h = v²/ (2g) = 3.78 / (2 x 9.80) = 0.19 m

Therefore, the maximum height reached by 1.40 kg box above the floor -

H = original 0.40 m + raised 0.40 m + h (0.19 m due to vertical velocity.)

= 0.99 m.