A worker pushing a 35.0-kg wooden crate at a constant speed for 13.7 m along a wood floor does 390 J of work applying a constant horizontal force of magnitude F0 on the crate.Determine the value for F0.
Work is applied force times distance.
F0 * 13.7 m = 390 J
F0 = 28.5 N
The speed does not matter. Neither does the weight.
A worker pushing a 35.0-kg wooden crate at a constant speed for 10.0 m along a wood
floor does 350 J of work by applying a constant horizontal force of magnitude F on the crate.
(a) Determine the value of F.
(b) Calculate the coefficient of kinetic friction.
(c) If the worker now applies a force greater than F, describe the subsequent motion of the
crate.
(d) If the work was done in 8 seconds, what is the average power?
To determine the value for F0, we need to use the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy.
In this case, since the crate is moving at a constant speed, its kinetic energy remains constant. Therefore, the work done by the worker is equal to zero.
From the problem, we know that the worker does 390 J of work on the crate. Since this work is done by applying a constant horizontal force, we can calculate the magnitude of the force using the formula:
Work = Force * Distance
Rearranging the formula, we have:
Force = Work / Distance
Plugging in the values, we get:
Force = 390 J / 13.7 m
Calculating this, we find that the value for F0 is approximately 28.47 N. So, the worker is applying a force of approximately 28.47 N to push the wooden crate.