Question

A 1.00-kg object is attached by a thread of negligible mass, which passes over a pulley of negligible mass, to a 2.00-kg object. The objects are positioned so that they are the same height from the floor and then released from rest. What are the speeds of the objects when they are separated vertically by 1.00 m?

Answer 1

Answer:

a = 3.27 m/s²

v = 2.56 m/s

Explanation:

given,

mass A = 1 kg

mass B = 2 kg

vertical distance between them = 1 m

$F_d = mg$

$F_d = 2 \times 9.8$

$F_d = 19.6\ N$

$F_u = mg$

$F_u = 1 \times 9.8$

$F_u = 9.8\ N$

$F_{net} = 19.6 - 9.8$

$F_{net}=9.8\ N$

$F = (m_1+m_2)a$

$9.8 = (2+1)a$

a = 3.27 m/s²

The speed of the system at that moment is:

v² = u² + 2×a×s

v² = 0² + 2× 3.27 × 1

v ² = 6.54

v = 2.56 m/s