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

Answer 2

The speeds of the objects when they are separated vertically by 1.00 m is mathematically given as

v = 2.56 m/s

What are the speeds of the objects when they are separated vertically by 1.00 m?

Question Parameters:

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 speeds of the objects when they are separated vertically by 1.00 m?

Generally the equation for the  Force is mathematically given as

F=mg

Therefore

F=2*9.8

Fd=19.6N

and

Fu=1*9.9

Fu=9.8

Hence the difference of the forces is

Fn=19.6N-9.8

Fn=9.8

Generally

F=(m1+m2)a

9.8 = (2+1)a

a = 3.27 m/s²

Generally the equation for the  Speed is mathematically given as

v^2 = u^2+ 2as

Hence

v^2 = 0^2 + 2* 3.27 * 1

v = 2.56 m/s

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