In a simple Atwood machine, two unequal masses $m_1$ and $m_2$ are connected by a string going over a clamped light smooth pulley. In an arrangement as shown in the figure, $m_1 = 300 \text{ g}$ and $m_2 = 600 \text{ g}$ . The system is released from rest. Determine the distance travelled by the first block in the first two seconds.

Question

In a simple Atwood machine, two unequal masses

m_1

and

m_2

are connected by a string going over a clamped light smooth pulley. In an arrangement as shown in the figure,

m_1 = 300 \text{ g}

and

m_2 = 600 \text{ g}

. The system is released from rest. Determine the distance travelled by the first block in the first two seconds.
In a simple Atwood machine, two unequal masses m_1 and m_2 are connected by a string going over a clamped light smooth

In a simple Atwood machine, two unequal masses m_1 and m_2 are connected by a string going over a clamped light smooth

Quantrex Academy · Accepted Answer

Correct answer: D. 6.5 m

The acceleration of the blocks in a simple Atwood machine is given by:

a = \dfrac{m_2 - m_1}{m_1 + m_2} g

Substituting the given values

m_1 = 300 \text{ g}

and

m_2 = 600 \text{ g}

:

a = \dfrac{600 - 300}{600 + 300} g = \dfrac{300}{900} g = \dfrac{g}{3}

Taking

g = 9.8 \text{ m/s}^2

:

a = \dfrac{9.8}{3} \text{ m/s}^2

The system is released from rest, so the initial velocity

u = 0

. The distance travelled by the first block in time

t = 2 \text{ s}

is:

s = ut + \dfrac{1}{2} a t^2

s = 0 + \dfrac{1}{2} \left(\dfrac{9.8}{3}\right) (2)^2

s = \dfrac{1}{2} \times \dfrac{9.8}{3} \times 4

s = \dfrac{19.6}{3} \approx 6.5 \text{ m}

In a simple Atwood machine, two unequal masses m_1 and m_2 are...

Options

Correct answer

Step-by-step solution

More from Newton's Laws of Motion