A 933 kg block is pushed on the slope of a 20
◦
frictionless inclined plane to give it an initial
speed of 50 cm/s along the slope when the
block is 1.5 m from the bottom of the incline.
The acceleration of gravity is 9.8 m/s
2
.
1.5 m
933 kg
20
◦
What is the speed of the block at the bottom of the plane?
Answer in units of m/s
To find the speed of the block at the bottom of the plane, we can use the concepts of conservation of energy. The block starts with a certain amount of gravitational potential energy at the top of the incline, and as it moves down the slope, this potential energy gets converted into kinetic energy.
First, let's find the initial potential energy of the block at the top of the incline. The potential energy is given by the formula:
Potential Energy = mass * gravity * height
where mass is the mass of the block, gravity is the acceleration due to gravity, and height is the vertical distance from the top of the incline to the block's position:
Potential Energy = 933 kg * 9.8 m/s^2 * 1.5 m
Next, let's find the final kinetic energy of the block at the bottom of the plane. The kinetic energy is given by the formula:
Kinetic Energy = (1/2) * mass * velocity^2
where mass is the mass of the block and velocity is the speed of the block at the bottom of the plane.
Since energy is conserved, the initial potential energy of the block is equal to the final kinetic energy of the block:
Potential Energy = Kinetic Energy
933 kg * 9.8 m/s^2 * 1.5 m = (1/2) * 933 kg * velocity^2
Now, we can solve this equation to find the velocity:
velocity^2 = (2 * 933 kg * 9.8 m/s^2 * 1.5 m) / 933 kg
velocity^2 = 29.4 m^2/s^2
velocity = sqrt(29.4 m^2/s^2)
velocity ≈ 5.42 m/s
Therefore, the speed of the block at the bottom of the plane is approximately 5.42 m/s.