A boy climbs up an 8.00m high tree. To come back down, the boy slides down along a rope. The coefficient of friction between the rope and the boy is 0.40. What is the velocity with which the boy reaches ground?
To find the velocity with which the boy reaches the ground, we can use the principle of conservation of mechanical energy.
First, let's calculate the potential energy of the boy when he is at the top of the tree. The potential energy (PE) is given by the equation PE = mgh, where m is the mass of the boy, g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the tree (8.00m). Since the mass of the boy is not given in the question, we can assume a value of 60 kg for an average boy.
PE = mgh = (60 kg)(9.8 m/s^2)(8.00m) = 4,704 J
Now, as the boy slides down, some of the potential energy is converted into kinetic energy. The remaining energy is lost due to friction between the rope and the boy. The work done by friction can be calculated using the equation W = μmgd, where μ is the coefficient of friction (0.40), m is the mass, g is the acceleration due to gravity, and d is the distance traveled.
Since the height of the tree is equal to the distance traveled along the rope, we can calculate the work done by friction using d = 8.00m.
W = μmgd = (0.40)(60 kg)(9.8 m/s^2)(8.00m) = 1,881.6 J
The work done by friction is equal to the energy lost by the boy during his descent. Therefore, the remaining energy is converted into kinetic energy.
Kinetic energy (KE) is given by the equation KE = (1/2)mv^2, where m is the mass, and v is the velocity.
KE = PE - W = 4,704 J - 1,881.6 J = 2,822.4 J
Now we can solve for the velocity:
2,822.4 J = (1/2)(60 kg)v^2
v^2 = (2 * 2,822.4 J) / (60 kg)
v^2 = 94.08 m^2/s^2
Taking the square root of both sides, we find:
v = √(94.08 m^2/s^2) ≈ 9.70 m/s
Therefore, the velocity with which the boy reaches the ground is approximately 9.70 m/s.
To find the velocity with which the boy reaches the ground, we can use the concept of conservation of energy.
The initial potential energy (mgh) the boy has when he is at the top of the tree will be converted into the final kinetic energy (0.5mv^2) when he reaches the ground.
Let's assume the mass of the boy is m, the height of the tree is h = 8.00m, and the coefficient of friction is μ = 0.40.
The potential energy at the top of the tree is given by:
PE = mgh
The work done by friction during the slide is given by:
W_friction = μmgd
The work done by friction will subtract energy, so we need to subtract it from the initial potential energy:
PE_final = PE_initial - W_friction
Since the final energy is all in the form of kinetic energy, we can set this equal to the kinetic energy (0.5mv^2):
0.5mv^2 = PE_initial - μmgd
Rearranging the equation, we get:
v^2 = (2gh - 2μgd)/m
Taking the square root of both sides, we get:
v = √((2gh - 2μgd)/m)
Now we can substitute the given values:
v = √((2 * 9.8m/s² * 8.00m - 2 * 0.40 * 9.8m/s² * 8.00m)/m)
Simplifying:
v = √((156.8m²/s² - 62.72m²/s²)/m)
v = √((94.08m²/s²)/m)
v = √(94.08m/s²)
So, the velocity with which the boy reaches the ground is approximately 9.70 m/s.