Question

Two blocks of masses 20 kg and 10 kg are connected by a mass-less pulley . How much time will the 20 kg block will take to travel a distance of 0.6 m?

Answer 1

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

$\large{m_1 = 20 \: Kg}$

$\large{m_2 = 10 \: Kg}$

d = 0.6 m.

$\huge{\bold{\underline{Explanation:-}}}$

We need to find first acceleration of the system.

$\large{\bold{a = \frac{(m_1 - m_2)g}{m_1 + m_2}}}$

Substituting the values.

$\large{ a = \frac{(20 - 10) \times 10}{20 + 10}}$

$\large{ a = \frac{10 \times \cancel{10}}{ \cancel{30}}}$

$\large{ a = \frac{10}{3} \: m/s^2}$

Now,applying second kinematic equation.

$\large{\bold{S = ut + \frac{1}{2}at^2}}$

In this case the initial velocity will be zero.

$\large{\bold{S = \frac{1}{2}at^2}}$

Substituting the values.

$\large{ 0.6 = \frac{1}{2} \times \frac{10}{3} \times t^2}$

$\large{t^2 = \frac{3.6}{10}}$

$\large{t^2 = \frac{36}{100}}$

$\large{t = \sqrt{ \frac{36}{100}}}$

$\large{t = \frac{6}{10}}$

$\huge{\boxed{\boxed{t = 0.6 \: Seconds}}}$

So, block will travel 0.6 m in 0.6 seconds.

Answer 2

$\huge{\underline{\underline{\mathfrak{Answer \colon}}}}$

From the Question,

Mass of the objects:

• M = 20Kg

• m = 10Kg

• Displacement,s = 0.6m

To find

Time taken to cover a distance of 0.6m

Acceleration of two masses attached by a string is given by:

$\boxed{ \boxed{ \sf{a = \frac{M - m}{M + m} }.g}}$

Putting the values,we get:

$\sf{a = \frac{20 - 10}{20 + 10}.10 } \\ \\ \implies \: \sf{a = \frac{10}{30}.10 } \\ \\ \implies \: \huge{\sf{a = 3.33 \: ms{}^{ - 2} }}$

Using the relation,

$\huge{ \sf{s = ut + \frac{1}{2}at{}^{2} }}$

Putting the values,we get:

$\sf{0.6 = \frac{1}{2} \times 3.33 .t {}^{2} } \\ \\ \implies \: \sf{t{}^{2} = \frac{1.2}{3.3} } \\ \\ \implies \: \huge{{\sf{t = 0.6s}}}$

• Thus,the object takes 0.6 seconds to cover a distance of 0.6m