Question

Two blocks of masses $m_1$ = 1 kg and $m_2$ = 4 kg connected by an ideal un-deformed spring rest on a horizontal plane. The coefficient of friction between the blocks and the surface is equal to $\mu = \frac{1}{5}$. What minimum constant force (in N) has to be applied in the horizontal direction to the block of mass $m_1$ in order to shift the other block? (Take g = 10 m/s²)

• A. 2
• B. 4
• C. 10
• D. 20

Answer 1

Answer: (C)

10

Answer 2

To find the minimum constant force $F$ required to shift block $m_2$, we need to analyze the forces acting on both blocks at the moment $m_2$ just begins to move.

1. Condition for block $m_2$ to start moving:

Block $m_2$ will start moving when the spring force ($F_s$) acting on it to the right overcomes the maximum static friction ($f_{s2,max}$) acting on it to the left.

The maximum static friction on $m_2$ is given by:

$f_{s2,max} = \mu N_2 = \mu m_2 g$

For $m_2$ to just start moving, its acceleration $a_2$ is approximately zero. Therefore, the net force on $m_2$ is zero:

$F_s - f_{s2,max} = m_2 a_2 = 0$

$F_s = f_{s2,max}$

$F_s = \mu m_2 g$

2. Forces on block $m_1$ at the threshold:

At the instant $m_2$ just starts moving, the spring has stretched by an amount such that it exerts a force $F_s = \mu m_2 g$ on $m_2$. This same spring force $F_s$ acts on $m_1$ to the left.

The applied force $F$ acts on $m_1$ to the right.

The maximum static friction ($f_{s1,max}$) acts on $m_1$ to the left.

$f_{s1,max} = \mu N_1 = \mu m_1 g$

To find the minimum constant force $F$, we assume that at this threshold, block $m_1$ is also on the verge of moving, meaning its acceleration $a_1$ is also approximately zero. If $m_1$ were accelerating, a larger force $F$ would be required.

The net force on $m_1$ is zero:

$F - F_s - f_{s1,max} = m_1 a_1 = 0$

$F = F_s + f_{s1,max}$

3. Substitute and Calculate:

Substitute the expressions for $F_s$ and $f_{s1,max}$:

$F = \mu m_2 g + \mu m_1 g$

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

Now, plug in the given values:

$m_1 = 1$ kg

$m_2 = 4$ kg

$\mu = \frac{1}{5}$

$g = 10$ m/s²

$F = \frac{1}{5} \times 10 \times (1 + 4)$

$F = 2 \times 5$

$F = 10$ N

The minimum constant force that has to be applied to block $m_1$ in order to shift block $m_2$ is 10 N.