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Answer:
v²/2gsinθ
The correct answer
This represents the distance traveled by the block up the inclined plane before coming to rest ..
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The correct answer is 2) (v^2)/(2g * sin theta).
When the block slides down the rough inclined plane with a constant velocity, the net force acting on the block is zero. This means that the frictional force (f) opposing the motion is equal in magnitude and opposite in direction to the component of the gravitational force (mg) along the plane, which is mg * sin(theta).
When the block is projected up the same plane with an initial velocity v, it will eventually come to rest when the net force acting on it is zero again. At the point of rest, the only force acting on the block is the component of the gravitational force along the plane (mg * sin(theta)), acting in the opposite direction to the motion.
Using the work-energy principle, we can equate the work done by the frictional force during the downward slide to the work done against gravity during the upward motion:
Frictional force * distance traveled down = mg * sin(theta) * distance traveled up
The distance traveled down the plane is equal to the distance traveled up the plane, so we can simplify this equation to:
f * s = mg * sin(theta) * s
Since the block is moving with a constant velocity down the plane, the work done by the frictional force is equal to zero (f * s = 0). Therefore, we are left with:
0 = mg * sin(theta) * s
Rearranging this equation, we find:
s = 0
This means that the block comes to rest at the same point it started from.
Now, let's consider the distance traveled up the plane before coming to rest. We can use the work-energy principle again:
Work done against gravity = Change in kinetic energy
The work done against gravity is given by the formula:
mg * sin(theta) * s = (1/2) * m * v^2
Dividing both sides by m and simplifying, we get:
g * sin(theta) * s = (1/2) * v^2
Rearranging this equation, we find:
s = (v^2) / (2g * sin(theta))
Therefore, the correct answer is 2) (v^2)/(2g * sin(theta)).