A 2 kg metal plate slides down a 11-meter high slope. At the bottom its speed is 7.5 m/s. To the nearest Joule, what was the magnitude of the work done by friction?If the slope in the above problem is 22 degrees, what is the coefficient of friction?
InitialPE-friction=final KE
solve for friction work.
Now, friction work= mu*mg*cosTheta*11meters
solve for mu.
To find the magnitude of the work done by friction, we need to calculate the change in mechanical energy of the metal plate.
First, we can calculate the potential energy of the plate at the top of the slope using the formula:
Potential Energy = mass × gravitational acceleration × height
Given:
Mass of the metal plate (m) = 2 kg
Height of the slope (h) = 11 meters
Gravitational acceleration (g) = 9.8 m/s^2 (approximate value on Earth)
Potential Energy = 2 kg × 9.8 m/s^2 × 11 meters
= 215.6 Joules (approximate value)
Next, we can calculate the final kinetic energy of the metal plate at the bottom of the slope using the formula:
Kinetic Energy = 0.5 × mass × speed^2
Given:
Mass of the metal plate (m) = 2 kg
Speed of the metal plate (v) = 7.5 m/s
Kinetic Energy = 0.5 × 2 kg × (7.5 m/s)^2
= 56.25 Joules
Since there is no loss or gain of energy due to any external force acting on the plate, the work done by friction can be determined by the change in mechanical energy:
Work by Friction = (Final Kinetic Energy) - (Initial Potential Energy)
Work by Friction = 56.25 Joules - 215.6 Joules
= -159.35 Joules
The magnitude of the work done by friction is approximately 159.35 Joules (rounded to the nearest Joule).
Now, to find the coefficient of friction, we can use the following formula:
Coefficient of Friction = tan(angle of the slope)
Given:
Angle of the slope = 22 degrees
Coefficient of Friction = tan(22 degrees)
= 0.404
Therefore, the coefficient of friction is approximately 0.404.