# Why use Work-Energy Theorem here?

My confusion is in reference to the solution of this problem, whose problem statement and its solution I restate here:

Problem:

Two blocks of masses $$m_1$$ and $$m_2$$ are placed on a rough horizontal surface (coefficient of friction $$\mu$$), connected by a light spring. Find the minimum constant force that has to be applied on the block with mass $$m_1$$ so that the other block just begins to slide.

Solution:

For second block $$μm_2g=kx.$$ For first block $$F\cdot x−\frac 12 kx^2−μm_1gx=\frac 12 m_1v^2.$$ Set $$v=0,$$ to get $$F=\frac 12 kx+μm_1g=\boxed {μg\left(m_1+\frac {m_2}2\right)}.$$

Here $$k$$ is the spring constant, $$\mu$$ is the coefficient of friction, $$x$$ is the change in length of spring and $$F$$ is the minimum external force applied on mass $$m_1$$.

Now, my confusion: Why can't this problem be solved using only Newton's Laws? Why is it so necessary to use the work-energy theorem here? Is it because of friction?

For mass $$m_1$$, $$F-f_1 = F_s, \qquad (1)$$ where $$f_1$$ is the frictional force on $$m_1$$, and $$F_s$$ is the spring force.

Now, we also have $$f_2 = \mu m_2 g = F_s, \qquad (2)$$ so combining $$(1)$$ and $$(2)$$ gives us $$F = \mu g(m_1 + m_2).$$

• Have you tried solving it using Newton's Laws of Motion only? – Eagle Mar 12 at 11:55
• Yes, but I couldn't get the right expression in the end. – Apekshik Panigrahi Mar 12 at 11:57
• Please add whatever you tried in the question. – Eagle Mar 12 at 11:57
• @Natasha I have now added the needed. – Apekshik Panigrahi Mar 12 at 12:29

Force required to just move the block with mass $$m_2$$ is $$F= \mu m_2g$$. Hence, the maximum extension $$(x_{max})$$ the spring can have is $$\mu m_2g/k$$.
According to Newton's second law, for $$m_1$$ we can write $$F_{net} = m_1a_1$$ $$F - kx - \mu m_1g = m_1a_1$$ where $$x$$ is the extension of spring.
We find, $$a_1 = \displaystyle \frac{F - kx - \mu m_1g}{m_1}$$.
We can express acceleration as $$a = v \displaystyle\frac{dv}{dx}$$, so $$(F - kx - \mu m_1g)dx = m_1 v dv$$ Upon integrating $$F\cdot x - \frac{1}{2} k x^2 - \mu m_1 g x = \frac{1}{2} m_1 v^2$$ we arrive at the same equation as given by work-energy theorem.