A block slides a distance d down a frictionless plane and then comes to a stop after sliding a distance s across a rough horizontal plane.
What fraction of the distance s does the block slide before its speed is reduced to one-third of the maximum speed it had at the bottom of the ramp?
A. s/3
B. 2s/3
C. s/9
D. 8s/9
8s/9
To solve this problem, let's consider the different stages of the block's motion.
1. Sliding down the frictionless plane:
When the block slides down the frictionless plane, its speed does not change, so its maximum speed is maintained throughout this distance.
2. Sliding across the rough horizontal plane:
When the block reaches the horizontal plane, the friction will gradually slow it down until it ultimately comes to a stop. The problem states that the block's speed is reduced to one-third of the maximum speed it had at the bottom of the ramp.
Now, let's calculate the fraction of the distance s the block slides before its speed is reduced to one-third:
Let's assume that the block slides a distance x before its speed is reduced.
From the problem statement, we know that the distance traveled on the horizontal plane (s) is equal to x + (2/3)x.
Simplifying the equation, we get:
s = x + (2/3)x
s = (5/3)x
Now, let's isolate x:
x = (3/5)s
Therefore, the fraction of the distance s that the block slides before its speed is reduced is x/s, which is (3/5)s divided by s, which simplifies to 3/5.
So, the answer is (A) s/3.
To determine the fraction of distance s that the block slides before its speed is reduced to one-third of the maximum speed it had at the bottom of the ramp, we need to analyze the situation and apply the relevant principles of physics.
Let's break it down step by step:
1. The initial speed of the block at the bottom of the ramp is given as the maximum speed. Let's call it v0.
2. As the block slides down the frictionless plane, its potential energy is converted to kinetic energy. According to the principle of conservation of energy, the potential energy lost is equal to the kinetic energy gained. This can be expressed as mgh = (1/2)mv0^2, where m is the mass of the block and g is the acceleration due to gravity.
3. At the bottom of the ramp, the block's speed is v0. We're interested in finding the distance fraction where its speed reduces to one-third of v0. So, the final speed we're looking for is (1/3)v0.
4. On the horizontal plane, the block encounters friction. The force of friction acts in the opposite direction to the block's motion, causing deceleration.
5. The work done by the force of friction on the block can be calculated using the work-energy principle. W = F * d = (1/2)mvf^2 - (1/2)mv0^2, where W is the work done by the frictional force, F is the magnitude of the frictional force, d is the distance traveled across the rough horizontal plane, and vf is the final speed of the block.
6. Since we're interested in the distance fraction s, where the speed reduces to (1/3)v0, we need to find vf in terms of s.
7. The distance s can be related to the time taken to reach point s across the rough horizontal plane using the equation s = (1/2)at^2, where a is the acceleration caused by the frictional force.
8. The acceleration a can be found by dividing the work done by the frictional force by the distance s: a = (2W) / (m * s).
9. Using the equation of motion, vf^2 = v0^2 - 2as, we can solve for vf in terms of s.
10. Once we find vf, we can equate it to (1/3)v0 and solve for the distance fraction s.
Now, let's apply the steps and determine the answer:
1. Start with the equation mgh = (1/2)mv0^2.
2. Solving for h, we get h = (1/2)v0^2 / g.
3. The work done by the frictional force is W = F * d = (1/2)mvf^2 - (1/2)mv0^2.
4. Substitute vf^2 = v0^2 - 2as into the work equation to get W = (1/2)m(v0^2 - 2as) - (1/2)mv0^2.
5. Simplify the equation to W = (1/2)mv0^2 - mas - (1/2)mv0^2.
6. The work done by the frictional force can also be written as W = F * s, where F is the magnitude of the frictional force.
7. Substitute a = (2W) / (m * s) into the equation to get a = (2W) / (m * s) = (mv0^2 - 2as) / (m * s).
8. Further simplify the equation to a = (v0^2 - 2as) / s.
9. Rearrange the equation to v0^2 - 2as = as.
10. Solve for vf in terms of s: vf = sqrt(v0^2 - as).
11. Equate vf to (1/3)v0: sqrt(v0^2 - as) = (1/3)v0.
12. Square both sides of the equation: v0^2 - as = (1/9)v0^2.
13. Solve for as: as = (8/9)v0^2.
14. Divide as by v0^2: as / v0^2 = 8/9.
15. Finally, since v0^2 was defined as the maximum speed at the bottom of the ramp, we can substitute s for v0^2 in the equation to get the fraction of distance s as as / s = 8/9.
Therefore, the fraction of the distance s that the block slides before its speed is reduced to one-third of the maximum speed it had at the bottom of the ramp is 8/9.
The correct answer is (D) 8s/9.