a 1-kilogram block slides down a frictionless inclined plane with an angle of 35 degrees and 1.0 meters high. what is its velocity at the bottom of the plane?
To find the velocity of the block at the bottom of the inclined plane, we can use the principle of conservation of energy.
The potential energy at the top of the inclined plane is given by:
Potential energy (PE) = mgh
where:
m = mass of the block = 1 kg
g = acceleration due to gravity = 9.8 m/s^2 (approximately)
h = height of the inclined plane = 1.0 m
PE = (1 kg)(9.8 m/s^2)(1.0 m)
PE = 9.8 J
At the bottom of the inclined plane, all of the potential energy has been converted to kinetic energy. Therefore, we can equate the potential energy at the top to the kinetic energy at the bottom:
Potential energy (PE) = Kinetic energy (KE)
KE = 1/2 mv^2
where:
v = velocity of the block at the bottom of the inclined plane (what we want to find)
9.8 J = 1/2 (1 kg) v^2
Rearranging the equation:
2 × 9.8 J = v^2
v^2 = 19.6 J
Taking the square root of both sides:
v = √19.6 J
v ≈ 4.427 m/s
Therefore, the velocity of the block at the bottom of the inclined plane is approximately 4.43 m/s.
To find the velocity of the block at the bottom of the inclined plane, we can use the principles of conservation of energy.
First, let's determine the potential energy of the block at the top and compare it to the kinetic energy at the bottom. The potential energy (PE) of an object is given by the formula: PE = mgh, where m is the mass of the object, g is the acceleration due to gravity (approximately 9.8 m/s²), and h is the height.
PE at the top = (1 kg) * (9.8 m/s²) * (1.0 m) = 9.8 Joules
According to the law of conservation of energy, the potential energy at the top will be converted entirely into kinetic energy at the bottom, neglecting other factors such as heat or friction. So, we can equate the potential energy at the top to the kinetic energy (KE) at the bottom:
PE at the top = KE at the bottom
9.8 Joules = (1/2) * m * v²
where v is the velocity of the block at the bottom of the inclined plane.
Now we can solve for v:
9.8 Joules = (1/2) * (1 kg) * v²
Multiplying both sides by 2 to eliminate the fraction:
19.6 Joules = v²
Taking the square root of both sides to solve for v:
v = √(19.6 Joules) ≈ 4.43 m/s
So, the velocity of the 1-kilogram block at the bottom of the frictionless inclined plane is approximately 4.43 m/s.
V^2 = Vo^2 + 2g*h
V^2 = 0 + 19.6*1 = 19.6
V = 4.43 m/s.