A worker pushes a 820 N crate with a horizontal force of 345 N a distance of 24 m. Assume the coefficient of kinetic frictin between the crate and the floor is 0.22.
a) How much work is done by the worker on the crate?
b) How much work is done by the floor?
c) What is the net work done on the crate?
a) 345 N*24 m = ___ J
b) -820 N*(0.22)*24 m = - ___ J
c) Add the previous two results.
(Note that (b) is negative)
To find the answers to these questions, we need to calculate the work done by different forces acting on the crate.
a) To determine the work done by the worker on the crate, we will use the formula:
Work = Force × Distance × Cos(θ)
Here, the force is the horizontal force the worker applies, which is 345 N, the distance is 24 m, and θ is the angle between the force and the displacement. Since the force is horizontal and the displacement is also horizontal, the angle θ is 0 degrees. In this case, Cos(θ) equals 1.
So, the work done by the worker can be calculated as:
Work = 345 N × 24 m × Cos(0°) = 8,280 J
Therefore, the worker does 8,280 Joules of work on the crate.
b) The work done by the floor can be determined using the equation:
Work = Force × Distance × Cos(θ)
Since the floor doesn't apply any horizontal force on the crate (only the worker does), there is no work done by the floor on the crate. Therefore, the answer is zero:
Work = 0 J
c) The net work done on the crate is the sum of the work done by all the forces acting on it. In this case, since there is no work done by the floor, the net work done on the crate is equal to the work done by the worker:
Net work = Work done by the worker = 8,280 J
Therefore, the net work done on the crate is 8,280 Joules.