In an electronics factory, small cabinets slide down a 30 degree incline a distance of 16 m along the incline to the next assembly stage. The cabinets have a mass of 10 kg each. Calculate the speed each cabinet would acquire if the incline were frictionless. What kinetic energy would a cabinet have?
To calculate the speed the cabinets would acquire if the incline were frictionless, you can use the principles of physics.
First, let's determine the gravitational force acting on the cabinet. The force due to gravity can be calculated using the formula:
F_gravity = m * g
where m represents the mass of the cabinet (10 kg) and g is the acceleration due to gravity (approximately 9.8 m/s^2).
F_gravity = 10 kg * 9.8 m/s^2 = 98 N
Next, we need to determine the component of the gravitational force parallel to the incline. This force can be found by multiplying the gravitational force by the sine of the incline angle:
F_parallel = F_gravity * sin(theta)
where theta represents the incline angle (30 degrees).
F_parallel = 98 N * sin(30 degrees) = 98 N * 0.5 = 49 N
The net force acting on the cabinet down the incline can be found by subtracting the force opposing the motion due to the incline:
F_net = F_parallel - Friction
Since the incline is assumed to be frictionless, the force opposing the motion is zero, so:
F_net = F_parallel = 49 N
Next, we can use Newton's second law of motion to find the acceleration of the cabinet down the incline:
a = F_net / m
a = 49 N / 10 kg = 4.9 m/s^2
Now, we can use the kinematic equation to find the final velocity (Vf) of the cabinet:
Vf^2 = Vi^2 + 2 * a * d
where Vi is the initial velocity (assumed to be zero), a is the acceleration (4.9 m/s^2), and d is the distance traveled (16 m).
Vf^2 = 0 + 2 * 4.9 m/s^2 * 16 m
Vf^2 = 156.8 m^2/s^2
Vf = sqrt(156.8) m/s
Vf ≈ 12.5 m/s
Therefore, if the incline were frictionless, each cabinet would acquire a speed of approximately 12.5 m/s.
To calculate the kinetic energy of the cabinet, we can use the formula:
Kinetic Energy = (1/2) * m * Vf^2
where m is the mass of the cabinet (10 kg) and Vf is the final velocity (12.5 m/s) we just calculated.
Kinetic Energy = (1/2) * 10 kg * (12.5 m/s)^2
Kinetic Energy = 0.5 * 10 kg * 156.25 m^2/s^2
Kinetic Energy ≈ 781.25 Joules
Therefore, the kinetic energy of a cabinet sliding down the incline would be approximately 781.25 Joules.