Question

Two masses 2 kg and 4 kg are connected at the two ends of light inextensible string passing over a frictionless pulley. If the masses are released, then what is the acceleration of block 2 kg? [Take g = 10 m/s2 ]
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Answer 1

Given,

Two masses 2 kg and 4 kg are connected at the two ends of light inextensible string passing over a frictionless pulley as shown in picture.

Now,

let the common acceleration of both objects be a and tension of string be T

For object of mass of 4 kg

4g>T

Therefore, resultant force will try move mass of 4kg in downward direction,

So,

4g - T = 4a .................................(i)

For object of mass of 2kg,

T>2g

Therefore, resultant force will try move mass of 2kg in upward direction,

T - 2g = 2a .................................(II)

Adding Equation (i) and (ii),

We get,

4g - 2g = 6a

2g = 6a

2*10 = 6a

6a = 20

a = 10/3

a = 3.33 m/sec²

Answer 2

$\huge\underline\mathfrak\red{ANSWER}$

$mass (1): = 2kg \\ mass(2) = 4kg \\ \\ let \: the \: acceleration \: of \: the \: 4kg \\ block \: be \: a \: in \: downward \: direction \\ \\ so \: acc \: to \: newton \: s \: law \: of \: motion \\ \\ m(1)g - t = m(1)a......eq(1) \\ \\ where \: t \: is \: the \: \: tension \: in \: the \: \\ string \\ \\ for \: mass(2) \: the \: acceleration \: will \\ be \: same \: in \: magnitude \: but \: opposite \\ in \: direction \\ \\ so \\ t - m(2)g = m(2)a.....eq(2) \\ \\ add \: both \: eq(1) \: and \: eq(2) \\ \\ (t - m(2)g) + (m(1)g - t) = a(m(1) + m(2)) \\ \\ m(1)g - m(2)g = a(m(2) + m(1)) \\ \\ we \: know \: that \\ \\ \: m(1) = 4kg \\ \: m(2) = 2kg \\ \\ put \: values \\ \\ 4g - 2g = a(4 + 2) \\ \\ 2g = 6a \\ \\ a = \frac{20}{6} \\ \\ a = 3.33m {sec}^{ - 1}$

GIVEN:-

STRING ARE INEXTENSIBLE. DUE TO WHICH THE LENGTH OF THE STRING REMAIN CONSTANT. NOW THE LENGTH IS CONSTANT SO EACH BLOCK TRAVEL EQUAL DISTANCE IS EQUAL INTERVAL OF TIME. DUE TO WHICH ITS ACCELERATIONS WILL BE SAME.

ANSWER BY....

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