Jin is sitting on top of a hemispherical, frictionless igloo of radius 2.40 meters. His friend pushes him, giving him an initial speed. Jin slides along the igloo and loses contact with it after he has traveled 1.60 meters along the surface. What was his initial speed?
To solve this problem, we can use the principles of conservation of energy.
First, let's calculate the potential energy when Jin is at the top of the igloo. The potential energy is given by:
PE = m * g * h,
where m is the mass of Jin, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height of the hemispherical igloo.
Since Jin is sitting on the top of the igloo, the height (h) is equal to the radius of the igloo (2.40 meters).
PE = m * g * h = m * g * 2.40
Next, let's find the kinetic energy when Jin is moving along the surface of the igloo. The kinetic energy is given by:
KE = (1/2) * m * v^2,
where v is the velocity of Jin.
When Jin loses contact with the igloo, all of his potential energy is converted into kinetic energy.
PE = KE,
m * g * h = (1/2) * m * v^2,
g * h = (1/2) * v^2,
v^2 = 2 * g * h,
v = sqrt(2 * g * h).
Now we can substitute the known values into the equation:
v = sqrt(2 * 9.8 * 2.40).
Calculating the value, we get:
v ≈ 9.47 m/s.
Therefore, Jin's initial speed was approximately 9.47 m/s.
To find Jin's initial speed, we can use the concept of conservation of energy. Initially, Jin is at rest on top of the hemispherical igloo, so his initial kinetic energy is zero. As he slides along the surface of the igloo, his potential energy is converted into kinetic energy.
Initially, Jin's potential energy is given by the formula U = mgh, where m is his mass, g is the acceleration due to gravity, and h is the height above the reference point (in this case, the height of the igloo). Since the igloo is frictionless, there is no work done against friction, so there is no energy loss in the form of heat.
At the position where Jin loses contact with the igloo, his potential energy becomes zero. Therefore, all his initial potential energy must have been converted into kinetic energy. Thus, we can equate the initial potential energy to the final kinetic energy:
mgh = (1/2)mv^2
where v is Jin's initial speed.
Since the mass cancels out from both sides of the equation, we can simplify it to:
gh = (1/2)v^2
Now, let's plug in the known values. The radius of the igloo is given as 2.40 meters, so the height above the igloo can be calculated using the Pythagorean theorem:
h = sqrt(r^2 - (r - d)^2)
where r is the radius of the igloo and d is the distance traveled by Jin along the surface of the igloo.
Plugging in the values, we get:
h = sqrt(2.40^2 - (2.40 - 1.60)^2)
= sqrt(2.40^2 - 0.80^2)
= sqrt(5.76 - 0.64)
= sqrt(5.12) ≈ 2.26 meters
Therefore, we have:
g * 2.26 = (1/2)v^2
Rearranging the equation to solve for v, we get:
v = sqrt(2 * g * 2.26)
The acceleration due to gravity is approximately 9.8 m/s^2, so plugging in the value, we have:
v ≈ sqrt(2 * 9.8 * 2.26)
≈ sqrt(43.04)
≈ 6.56 m/s
Therefore, Jin's initial speed was approximately 6.56 m/s.