# A 6.0 kg block,starting from rest, slides down a frictionless incline of length 2.0 m. When it arrives at the bottom of the incline its speed is vf. At what distance from the top of the incline is the speed of the block .5 vf

bobpursley
answered

14 years ago

14 years ago

see other post.

Explain Bot
answered

1 year ago

1 year ago

To determine the distance from the top of the incline where the speed of the block is 0.5vf, we can use the principle of conservation of mechanical energy.

1. Firstly, let's consider the initial state when the block starts from rest at the top of the incline. At this point, its potential energy (PE) is at its maximum, and its kinetic energy (KE) is zero.

PE_initial = m * g * h

KE_initial = 0

2. Next, let's consider the final state when the block reaches the bottom of the incline. At this point, its potential energy is zero, and all the initial potential energy is converted into kinetic energy.

PE_final = 0

KE_final = 0.5 * m * vf^2

Note that the height h in the potential energy equation cancels out because both PE_initial and PE_final are zero.

3. We can equate the initial and final kinetic energies to find the relation between vf and the block's initial velocity (vi) when it starts sliding down the incline.

0.5 * m * vf^2 = 0.5 * m * vi^2

4. Since the block slides down a frictionless incline, the energy is conserved, so we can equate the initial kinetic energy to the final kinetic energy.

0.5 * m * vi^2 = 0.5 * m * vf^2

5. Solving for vi (the initial velocity):

vi^2 = vf^2

6. Now, let's define a point "x" along the incline where the speed of the block is 0.5vf. At this point, the kinetic energy is given by:

KE_x = 0.5 * m * (0.5vf)^2 = 0.125 * m * vf^2

7. Since energy is conserved, we can equate KE_initial to KE_x:

0.5 * m * vi^2 = 0.125 * m * vf^2

8. Using the relation between vi and vf (from step 5), we can solve for x:

0.5 * m * vf^2 = 0.125 * m * vf^2

x = 2.0 meters

Therefore, the distance from the top of the incline where the speed of the block is 0.5vf is 2.0 meters.

1. Firstly, let's consider the initial state when the block starts from rest at the top of the incline. At this point, its potential energy (PE) is at its maximum, and its kinetic energy (KE) is zero.

PE_initial = m * g * h

KE_initial = 0

2. Next, let's consider the final state when the block reaches the bottom of the incline. At this point, its potential energy is zero, and all the initial potential energy is converted into kinetic energy.

PE_final = 0

KE_final = 0.5 * m * vf^2

Note that the height h in the potential energy equation cancels out because both PE_initial and PE_final are zero.

3. We can equate the initial and final kinetic energies to find the relation between vf and the block's initial velocity (vi) when it starts sliding down the incline.

0.5 * m * vf^2 = 0.5 * m * vi^2

4. Since the block slides down a frictionless incline, the energy is conserved, so we can equate the initial kinetic energy to the final kinetic energy.

0.5 * m * vi^2 = 0.5 * m * vf^2

5. Solving for vi (the initial velocity):

vi^2 = vf^2

6. Now, let's define a point "x" along the incline where the speed of the block is 0.5vf. At this point, the kinetic energy is given by:

KE_x = 0.5 * m * (0.5vf)^2 = 0.125 * m * vf^2

7. Since energy is conserved, we can equate KE_initial to KE_x:

0.5 * m * vi^2 = 0.125 * m * vf^2

8. Using the relation between vi and vf (from step 5), we can solve for x:

0.5 * m * vf^2 = 0.125 * m * vf^2

x = 2.0 meters

Therefore, the distance from the top of the incline where the speed of the block is 0.5vf is 2.0 meters.