A worker pushes a 1.50*10^3 N crate with a horizontal force of 345 N a distance of 24.0 m. Assume the coefficient of kinetic friction between the crate and the floor is 0.220.
a. How much work is done by the worker on the crate?
b. How much work is done by the floor on the crate?
c. What is the net work done on the crate?
Getting a seems easy enough I guess multiplying 345 and 24.0. Still I don't understand the others, someone please help.
a. To calculate the work done by the worker on the crate, you can use the formula:
Work = Force x Distance x cos(angle)
In this case, the force applied by the worker is 345 N and the distance is 24.0 m. Since the force and displacement are both along the horizontal direction, the angle between them is 0 degrees (cos(0) = 1). Therefore, the work done by the worker is:
Work = 345 N x 24.0 m x cos(0°)
Work = 8280 Joules (or 8280 J)
b. To calculate the work done by the floor on the crate, you need to consider the work done by the friction force. The friction force can be calculated using the equation:
Friction Force = coefficient of friction x Normal Force
The normal force is equal to the weight of the crate, which is given as 1.50 * 10^3 N. So the friction force can be calculated as:
Friction Force = 0.220 x 1.50 * 10^3 N
Friction Force = 330 N
Since the crate is moving horizontally, the floor exerts a force equal in magnitude and opposite in direction to the friction force. Therefore, the work done by the floor is calculated using the same formula as in part (a) but with the friction force and the distance as input:
Work = Friction Force x Distance x cos(angle)
Since the force and displacement are in the opposite direction, the angle between them is 180 degrees (cos(180°) = -1). Therefore, the work done by the floor is:
Work = -330 N x 24.0 m x cos(180°)
Work = -7920 J
Note that the negative sign indicates that the work done by the floor is in the opposite direction to the displacement of the crate.
c. The net work done on the crate is the sum of the work done by the worker (positive work) and the work done by the floor (negative work):
Net Work = Work by worker + Work by floor
Net Work = 8280 J + (-7920 J)
Net Work = 360 J
So, the net work done on the crate is 360 Joules.
To calculate the work done by the worker on the crate, you can use the formula:
Work = Force x Distance
In this case, the force applied by the worker is 345 N, and the distance the crate is moved is 24.0 m. Multiplying these values together will give you the work done by the worker on the crate:
Work = 345 N x 24.0 m = 8,280 J
Now let's move on to the other parts.
b. To find the work done by the floor on the crate, we need to consider the force of friction acting opposite to the direction of motion. The force of kinetic friction can be calculated using the formula:
Force of Friction = Coefficient of Kinetic Friction x Normal Force
The normal force is the force exerted by the floor on the crate, which is equal to the weight of the crate (mg) since the crate is on a horizontal surface. The weight of the crate can be calculated using the formula:
Weight = Mass x Gravity
Given that the weight of the crate is 1.50 × 10^3 N and the coefficient of kinetic friction is 0.220, we can calculate the force of friction as:
Force of Friction = 0.220 x (1.50 × 10^3 N) = 330 N
Now, since the crate is moving horizontally, the force of friction opposes its motion. Therefore, the work done by the floor on the crate is negative because the displacement and force are in opposite directions. Thus, the work done by the floor on the crate is:
Work = -Force of Friction x Distance
Work = -330 N x 24.0 m = -7,920 J
c. The net work done on the crate can be found by subtracting the work done by the floor on the crate from the work done by the worker on the crate:
Net Work = Work done by the worker - Work done by the floor
Net Work = (8,280 J) - (- 7,920 J) = 16,200 J
Therefore, the net work done on the crate is 16,200 J.