Question

Two masses are
hanging from
either side of a
massless,
frictionless
pulley. The first
mass is 5 kg and
the second mass
is 6 kg
Determine the
acceleration of
the system when
the masses are
released.
0.89 m/s2
O 1.96m/s2
1.63m/s2
1.89m/s2​

Answer 1

Answer- The above question is from the chapter 'Laws of Motion'.

Concept used: If two blocks of masses m and M are connected by light inextensible string passing over a smooth fixed pulley of negligible mass such that M > m,

Forces acting on M block:-

1) Weight, Mg vertically downwards

2) Tension, T vertically upwards

Forces acting on m block:-

1) Weight, mg vertically downwards

2) Tension, T vertically upwards

Let their acceleration be 'a'.

The equations are:-

Mg - T = Ma --- (1)

T - mg = ma --- (2)

Adding both equations, we get,

Mg - mg = Ma + ma

(M - m)g = (M + m)a

a = (M - m)g/(M + m)

Put value of a in equation 1, we get,

Mg - T = M × (M - m)g/(M + m)

T = Mg - M × (M - m)g/(M + m)

T = Mg (M + m - M + m)(M + m)

T = 2Mmg/(M + m)

Given question: Two masses are hanging from either side of a massless, friction less pulley. The first mass is 5 kg and the second mass is 6 kg. Determine the acceleration of the system when the masses are

released.

a) 0.89 m/s²

b) 1.96 m/s²

c) 1.63 m/s²

d) 1.89 m/s²

Answer: Let mass of first block (m) = 5 kg

Mass of second block (M) = 6 kg

These two masses are hanging from either side of a massless, friction less

pulley.

Let their acceleration be denoted by 'a'.

We know that acceleration is given by

$\boxed{a = \dfrac{(M - m)g}{M + m}}$

Substituting the values, we get,

$a = \dfrac{(6 - 5) \times 9.8}{5 + 6}\\\\a = \dfrac{9.8}{11}\\\\\bold {\boxed{a = 0.89 \ m/s^{2}}}$

∴ a) 0.89 m/s² is the correct option.

Answer 2

Answer- The above question is from the chapter 'Laws of Motion'.

Concept used: If two blocks of masses m and M are connected by light inextensible string passing over a smooth fixed pulley of negligible mass such that M > m,

Forces acting on M block:-

1) Weight, Mg vertically downwards

2) Tension, T vertically upwards

Forces acting on m block:-

1) Weight, mg vertically downwards

2) Tension, T vertically upwards

Let their acceleration be 'a'.

The equations are:-

Mg - T = Ma --- (1)

T - mg = ma --- (2)

Adding both equations, we get,

Mg - mg = Ma + ma

(M - m)g = (M + m)a

a = (M - m)g/(M + m)

Put value of a in equation 1, we get,

Mg - T = M × (M - m)g/(M + m)

T = Mg - M × (M - m)g/(M + m)

T = Mg (M + m - M + m)(M + m)

T = 2Mmg/(M + m)

Given question: Two masses are hanging from either side of a massless, friction less pulley. The first mass is 5 kg and the second mass is 6 kg. Determine the acceleration of the system when the masses are

released.

a) 0.89 m/s²

b) 1.96 m/s²

c) 1.63 m/s²

d) 1.89 m/s²

Answer: Let mass of first block (m) = 5 kg

Mass of second block (M) = 6 kg

These two masses are hanging from either side of a massless, friction less

pulley.

Let their acceleration be denoted by 'a'.

We know that acceleration is given by

\boxed{a = \dfrac{(M - m)g}{M + m}}

a=

M+m

(M−m)g

Substituting the values, we get,

\begin{gathered}a = \dfrac{(6 - 5) \times 9.8}{5 + 6}\\\\a = \dfrac{9.8}{11}\\\\\bold {\boxed{a = 0.89 \ m/s^{2}}}\end{gathered}

a=

5+6

(6−5)×9.8

a=

11

9.8

a=0.89 m/s

2

∴ a) 0.89 m/s² is the correct option.