Our tutors scan the Jiskha board looking for subjects in which they have some knowledge.
As far as I know, none of our tutors know anything about uos.
As far as I know, none of our tutors know anything about uos.
university optional science?
university optimal services?
unrealized optical scans?
If it slides down a vertical distance of .4 meters then we can calculate the kinetic energy
(1/2) m v^2 = m g h
(1/2)(2)v^2 = 20 (.4) = 8 Joules, not 60 joules.
Moreover you asked for nothing, just stated impossible facts.
The total mechanical energy is the sum of potential energy and kinetic energy:
Total mechanical energy (E) = Potential energy (PE) + Kinetic energy (KE)
Given that the object starts from rest at the top of the inclined plane, it has no initial kinetic energy. Therefore, the total mechanical energy is equal to the potential energy at the top.
To find the potential energy, we can use the formula:
Potential energy (PE) = mgh
Where:
m = mass of the object (2.0 kg)
g = acceleration due to gravity (10 m/s^2)
h = height of the incline (0.40 m)
Substituting the values into the equation, we have:
PE = 2.0 kg * 10 m/s^2 * 0.40 m
PE = 8 J
Since the object has a kinetic energy of 60 J at the bottom, the total mechanical energy at the bottom is:
Total mechanical energy (E) = PE + KE
E = 8 J + 60 J
E = 68 J
Now, we can use the conservation of energy principle to find the speed at the bottom. The total mechanical energy at the top is equal to the kinetic energy at the bottom.
Therefore, we can equate the two:
E = KE
68 J = 1/2 * m * v^2
Remembering that the object has a mass of 2.0 kg, we can solve for the velocity (v):
68 J = 1/2 * 2.0 kg * v^2
Dividing both sides by 2.0 kg:
34 J/kg = v^2
Taking the square root of both sides:
v = √(34 J/kg)
v ≈ 5.83 m/s
Therefore, the speed of the object at the bottom of the inclined plane is approximately 5.83 m/s.