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Step-by-Step Solution
Step 1: Understand the situation when the mass slides down
When the body (mass $m = 2\text{ kg}$) slides down an inclined plane of angle $\theta = 30^\circ$ with an acceleration $a = 3\text{ m/s}^2$, the forces acting along the plane are:
• Component of gravitational force down the plane: $mg \sin \theta$
• Frictional force opposing the motion up the plane (let friction coefficient be $\mu$)
We are given that the resulting net acceleration is $3\text{ m/s}^2$.
Step 2: Express the equation of motion when sliding down
The net force acting down the incline is:
$$
mg \sin \theta - \mu \, mg \cos \theta
$$
According to Newton’s second law ($F = ma$), we have:
$$
mg \sin \theta - \mu \, mg \cos \theta = ma.
$$
Substituting $a = 3\text{ m/s}^2$, $g = 10\text{ m/s}^2$, $\theta = 30^\circ$ gives an equation that relates $\mu$ and $a$, but for this problem, the exact value of $\mu$ is not directly needed once we see the final expression for the force driving the body up.
Step 3: Express the equation of motion when the mass is pushed up
Now, an external force $F$ is applied to move the body up the incline with the same acceleration $a$ (i.e., $3\text{ m/s}^2$). In that case, the forces along the plane are:
• $F$ upward along the plane
• $mg \sin \theta$ downward along the plane
• Frictional force $\mu \, mg \cos \theta$ acting downward along the plane (opposing the motion)
Again applying Newton’s second law, the net upward force must be equal to $ma$:
$$
F - (mg \sin \theta + \mu \, mg \cos \theta) = ma.
$$
Hence,
$$
F = ma + mg \sin \theta + \mu \, mg \cos \theta.
$$
Step 4: Use the known relation from the downward motion
From the downward scenario, we know:
$$
a = g(\sin \theta - \mu \cos \theta).
$$
Therefore,
$$
\mu \, mg \cos \theta = mg \sin \theta - ma.
$$
Substituting this into the expression for $F$:
$$
F = ma + mg \sin \theta + \bigl(mg \sin \theta - ma\bigr)
= 2\, mg \sin \theta.
$$
Notice that the $\mu$ terms and $ma$ terms effectively combine to give $2\,mg \sin \theta$.
Step 5: Calculate the final numerical value
Substitute $m = 2\text{ kg}$, $g = 10\text{ m/s}^2$, and $\sin 30^\circ = \tfrac{1}{2}$:
$$
F = 2 \times 2 \times 10 \times \frac{1}{2} = 20\text{ N}.
$$
Therefore, the external force required to push the mass up the plane with the same acceleration of $3\text{ m/s}^2$ is
20 N.
