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


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.


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?

EDIT: My approach was this:

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 $$ F = \mu g(m_1 + m_2).$$

  • $\begingroup$ Have you tried solving it using Newton's Laws of Motion only? $\endgroup$
    – Eagle
    Commented Mar 12, 2019 at 11:55
  • $\begingroup$ Yes, but I couldn't get the right expression in the end. $\endgroup$ Commented Mar 12, 2019 at 11:57
  • $\begingroup$ Please add whatever you tried in the question. $\endgroup$
    – Eagle
    Commented Mar 12, 2019 at 11:57
  • $\begingroup$ @Natasha I have now added the needed. $\endgroup$ Commented Mar 12, 2019 at 12:29

1 Answer 1


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.

In essence, we don't need to use work-energy theorem per se, but it is very convenient.

  • $\begingroup$ So is it necessary to reach the work-energy theorem and then solve the system, or is there another way where this result can be avoided and yet would lead to a solution? $\endgroup$ Commented Mar 12, 2019 at 12:11

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