# A mass, m_1 of 2.7 kilograms block slides on a friction-less horizontal surface and is connected...

## Question:

A mass, m{eq}_1 {/eq} = 2.7 kg block slides on a frictionless horizontal surface and is connected on one side to a spring (k = 40 N/m). The other side is connected to the block m{eq}_2 {/eq} = 3.6 kg that hangs vertically. The system starts from rest with the spring unextended.

a) What is the maximum extension of the spring?

b) What is the speed of block m{eq}_2 {/eq} when the extension is 65 cm?

## Energy Conservation Principle:

The energy conservation principle consists of a physical law that states that for a closed system, the total energy of the system should conserve. Energy is not created nor destroyed, it only transforms. Energy transformations may be present inside the system, but the overall total energy of the system will remain the same. The total energy of a system can be determined by the sum of the total kinetic and potential energy of the system.

$$E_t=K_e+U_e\\ E_t=0.5mv^2+mgh$$

#### Part a):

To determine the extension in the spring, we must determine how energy conserves in the system. Initially, the hanging block will have energy stored as potential energy. As the block displaces down, the spring will stretch the same distance the system displaces as everything is joined together. We define the following relationship.

{eq}E_i=E_f\\ U_e=U_k\\ m_2gh=\frac{1}{2}kx^2\\ h=x\\ m_2g=0.5kx\\ x=\dfrac{2m_2g}{k}\\ x=\dfrac{2\times3.6\times9.8}{40}\\ x=1.76\ m {/eq}

#### Part b):

To calculate the speed of the block as it falls, we apply the conservation of energy principle to the point where the spring has stretched 63 cms.

{eq}E_i=E_f\\ 0=K_e+U_k+U_e\\ 0=\dfrac{1}{2}m_2v^2+\dfrac{1}{2}kx^2+m_2gx\\ 0=0.5\times 3.6v^2+0.5\times40\times 0.65^2+3.6\times9.8\times0.65\\ 0=1.8v^2+8.45+22.93\\ v=-\sqrt{\dfrac{31.38}{1.8}}\\ v=-4.18\ m/s {/eq}