A 2.7 kg block is moved at constant speed over a surface for which the coefficient of kinetic friction is 0.19. The displacement is 5 m. It is pushed by a force directed at 21 degrees below the horizontal as shown in the figure above.
Find the work done on the block by:
a) the force;
a) friction;
To calculate the work done on the block by a force, we use the formula:
Work = Force x Distance x cos(theta)
Where:
- Force is the magnitude of the force applied on the block
- Distance is the displacement of the block
- theta is the angle between the direction of the force and the displacement
a) The force applied on the block is given by the equation:
Force = m * g * sin(theta)
Where:
- m is the mass of the block (2.7 kg)
- g is the acceleration due to gravity (9.8 m/s^2)
- theta is the angle between the force direction and the horizontal (21 degrees below the horizontal)
Therefore, the force applied on the block is:
Force = 2.7 kg * 9.8 m/s^2 * sin(21 degrees)
Using a calculator, we find that the force is approximately 11.9 N.
Now, we can calculate the work done by the force:
Work = 11.9 N * 5 m * cos(21 degrees)
Using a calculator, we find that the work done by the force is approximately 56.2 J.
b) The work done by friction can be calculated using the formula:
Work = force of friction x distance
The force of friction can be found using the equation:
force of friction = coefficient of kinetic friction * normal force
The normal force is equal to the weight of the block, which can be calculated using the equation:
normal force = mass * gravity
The coefficient of kinetic friction is given as 0.19.
Therefore, the force of friction is:
force of friction = 0.19 * (2.7 kg * 9.8 m/s^2)
Using a calculator, we find that the force of friction is approximately 5.36 N.
Now, we can calculate the work done by friction:
Work = 5.36 N * 5 m
Using a calculator, we find that the work done by friction is approximately 26.8 J.