A uniform plank of length L leans against a vertical wall, its head a distance h above the horizontal ground. If we let go of the plank from rest, how far above the ground is the head of the plank the moment it loses contact with the wall?
To calculate the distance above the ground at the moment the plank loses contact with the wall, we need to consider the conservation of energy.
1. Start by identifying the potential and kinetic energy involved in the system:
- Potential energy: The potential energy depends on the height above the ground, given by mgh, where m is the mass of the plank, g is the acceleration due to gravity, and h is the distance the head is above the ground.
- Kinetic energy: Initially, the plank is at rest, so its kinetic energy is zero when it loses contact with the wall.
2. As the plank falls, potential energy is converted into kinetic energy. When the plank loses contact with the wall, all the potential energy is converted into kinetic energy.
3. Equate the initial potential energy to the final kinetic energy:
mgh = 0.5mv^2, where v is the velocity of the plank at the moment it loses contact with the wall.
4. Simplify the equation by canceling out the mass:
gh = 0.5v^2
5. Rearrange the equation to solve for the velocity:
v^2 = 2gh
6. Take the square root of both sides to find the velocity:
v = √(2gh)
7. Now, let's find the height above the ground at the moment the plank loses contact with the wall. Since the velocity is zero at this moment, we can use the equation of motion:
h = 0.5gt^2, where t is the time it takes for the plank to fall.
8. Solve for t by rearranging the equation:
t = √(2h/g)
9. Finally, substitute the value of t into the equation to find the height h:
h = 0.5g(√(2h/g))^2
10. Simplify the equation:
h = 2h
11. Solve for h:
h = 0
Therefore, the height above the ground at the moment the plank loses contact with the wall is zero.