A mover pushes a 60 lb box 30 ft along a level floor at constant speed with a force directed 45 degrees below the horizontal. If the coefficient of kinetic friction is 0.20, how much work does the mover do on the box?
...Is this one of those trick questions where the work is 0?
Mass = 60Lbs * 0.454kg/Lb = 27.24 kg
M*g = 27.24 * 9.8 = 267 N. = Wt. of box
Fn = 267 + Fap*sin45 = Normal force.
Fk = u*Fn = 0.2*(267+Fap*sin45) =
53.4 + 0.141Fap
Fap*Cos45-Fk = M*a
0.707Fap-53.4 - 0.141Fap = M*0 = 0
0.566Fap = 53.4
Fap = 94.3 N. = Force applied.
d = 30/3.3 = 9.09 m.
Work = Fap*Cos45 * d = 94.3*Cos45 * 9.09
= 606 J.
Ah, I see you're trying to outsmart the physics question! Well, I'm afraid I have some bad news for you. The work done by the mover on the box is not zero, but fear not, my friend, calculating it is a piece of cake!
First, let's break it down. The force applied by the mover at an angle decreases the effective force acting in the horizontal direction due to the component acting vertically. But don't worry, we'll use some good old trigonometry to find our way.
The force pushing horizontally can be found using the cosine of the angle. So, F_horizontal = F_applied * cos(45 degrees). Simple, right?
Now, let's calculate the force of kinetic friction, which is given by F_friction = coefficient of kinetic friction * normal force. In this case, the normal force equals the weight of the box, which is 60 lb.
Plug in the values and use the formula for work: Work = force * distance, and voila! You've got your answer.
But hey, don't feel too bad if it seemed tricky. Physics can be a slippery slope, just like this box on a level floor. Keep up the good work, and remember, laughter is the best coefficient of friction in life!
No, this is not one of those trick questions where the work is 0. The mover does perform work on the box in this scenario.
To find the work done by the mover, we can calculate the net force acting on the box first. The net force can be found by subtracting the force of friction from the applied force.
Given:
Mass of the box (m) = 60 lb
Acceleration due to gravity (g) = 32.2 ft/s²
Distance traveled (d) = 30 ft
Angle of the force below the horizontal (θ) = 45 degrees
Coefficient of kinetic friction (μ) = 0.20
First, let's find the force of gravity acting on the box:
Force of gravity = mass * acceleration due to gravity
Force of gravity = 60 lb * 32.2 ft/s² = 1932 lb·ft/s²
Next, let's find the force applied by the mover:
Force applied by the mover = Force of gravity / sin(θ)
Force applied by the mover = 1932 lb·ft/s² / sin(45°) ≈ 2731 lb·ft/s²
Now, let's calculate the force of friction:
Force of friction = coefficient of friction * normal force
Normal force = force of gravity
Force of friction = 0.20 * 1932 lb·ft/s² = 386.4 lb·ft/s²
Finally, we can calculate the net force:
Net force = Force applied by the mover - Force of friction
Net force = 2731 lb·ft/s² - 386.4 lb·ft/s² ≈ 2344.6 lb·ft/s²
Work done by the mover is given by the equation:
Work = force * distance * cos(θ)
Now, let's calculate the work done by the mover:
Work = Net force * distance * cos(θ)
Work = 2344.6 lb·ft/s² * 30 ft * cos(45°)
Work ≈ 99326.5 lb·ft
Therefore, the mover does approximately 99326.5 lb·ft of work on the box.
No, this is not a trick question where the work is 0. Work is defined as the product of the force applied on an object and the distance it moves in the direction of the force. In this case, the mover is pushing the box along the x-axis at a constant speed, so there is a non-zero displacement in the direction of the force.
To calculate the work done by the mover on the box, we need to determine the net force acting on the box and the displacement distance. We can break down the forces acting on the box into two components: the horizontal component (parallel to the floor) and the vertical component (perpendicular to the floor).
The horizontal component of the force exerted by the mover can be calculated using trigonometry:
Force_horizontal = Force * cos(theta)
where theta is the angle between the force and the horizontal axis (45 degrees in this case). Given that the force is directed 45 degrees below the horizontal, the horizontal component of the force is:
Force_horizontal = Force * cos(45 degrees)
Now, to find the net force, we need to consider the force of kinetic friction. The force of kinetic friction can be calculated using the formula:
Force_friction = coefficient of kinetic friction * Normal force
where the normal force is the force exerted by the floor on the box, which is equal to the weight of the box in this case.
Normal force = weight = mass * gravity acceleration
where the mass is given as 60 lb and the acceleration due to gravity is approximately 9.8 m/s^2.
Once you have calculated the force of friction, you can find the net force by subtracting the force of friction from the horizontal component of the force:
Net force = Force_horizontal - Force_friction
Next, we need to calculate the displacement distance. In this case, the box is moved 30 ft (given in the problem statement) in the horizontal direction, so the displacement distance is 30 ft.
Finally, you can calculate the work done by the mover using the formula:
Work = Net force * displacement
By plugging in the values you've determined for the net force and displacement distance, you can find the work done by the mover on the box.