To determine how far up a 14° incline a solid sphere can roll, we need to consider the conservation of mechanical energy. The initial kinetic energy of the sphere is converted into potential energy as it rolls up the incline.
The potential energy gained by the sphere is given by:
ΔPE = m * g * h
where:
ΔPE is the change in potential energy,
m is the mass of the sphere,
g is the acceleration due to gravity (approximately 9.8 m/s²),
h is the height gained by the sphere.
The initial kinetic energy of the sphere is given by:
KE = (1/2) * m * v^2
where:
KE is the kinetic energy of the sphere,
v is the speed of the sphere.
Since the sphere is rolling without slipping, the linear velocity is related to the angular velocity by:
v = ω * r
where:
ω is the angular velocity of the sphere,
r is the radius of the sphere.
For a solid sphere rolling without slipping, the relationship between the angular velocity and the linear velocity is:
ω = v / r
Combining the equations, we can express the kinetic energy in terms of the angular velocity:
KE = (1/2) * m * (v/r)^2
We can equate the initial kinetic energy to the change in potential energy:
(1/2) * m * (v/r)^2 = m * g * h
Simplifying the equation:
(v/r)^2 = 2 * g * h
Now, we can solve for h:
h = [(v/r)^2] / (2 * g)
Given:
v = 5.1 m/s (speed of the sphere)
r = radius of the sphere (which is not provided)
Unfortunately, without the radius of the sphere, we cannot calculate the exact height it can roll up the incline. The height gained by the sphere depends on the radius, as it affects the relationship between the linear and angular velocities.
If you have the radius of the sphere, please provide it so that I can calculate the height it can roll up the incline.