A block of mass 2kg slides down a plane having an inclination of ƒá=15.0„a . Ignore friction. The block starts from rest at the top of the incline and the length of the incline is 2.00m.
The velocity of the block when it reaches the bottom of the incline is?
(1/2) m v^2 = m g h
so
v =sqrt(2gh)
here h = 2 sin 15
3.2 m/s
To find the velocity of the block when it reaches the bottom of the incline, we can use the principles of energy conservation.
Step 1: Determine the potential energy at the top of the incline.
The potential energy of an object is given by the equation: PE = mgh, where m is the mass of the object, g is the acceleration due to gravity, and h is the height or vertical distance.
In this case, the block starts from rest at the top of the incline. Since it is at rest, the initial kinetic energy is zero. Therefore, all the potential energy is converted into kinetic energy as the block slides down. We can calculate the potential energy by using the equation: PE = mgh.
Given:
mass (m) = 2 kg
acceleration due to gravity (g) = 9.8 m/s^2
height (h) = 2 m (length of the incline)
PE = (2 kg) * (9.8 m/s^2) * (2 m)
PE = 39.2 Joules
Step 2: Determine the kinetic energy at the bottom of the incline.
The kinetic energy of an object is given by the equation: KE = 1/2 * m * v^2, where KE is the kinetic energy, m is the mass of the object, and v is the velocity.
Since the block starts from rest at the top of the incline, the initial kinetic energy is zero. Therefore, all the potential energy is converted into kinetic energy as the block slides down the incline. We can equate the potential energy to the kinetic energy equation:
PE = KE
39.2 Joules = 1/2 * (2 kg) * v^2
Simplifying the equation:
78.4 J = v^2
Taking the square root of both sides:
v = √78.4
Calculating the square root:
v ≈ 8.85 m/s
Therefore, the velocity of the block when it reaches the bottom of the incline is approximately 8.85 m/s.