Calculating Work Done for a Box on Horizontal Floor

[chawki]

Homework Statement

A box, whose weight is 182 kg, is drawn with constant speed 21 m along a horizontal concrete floor. The kinetic friction between the box and the floor is 0.58.
(acceleration due to gravity, g = 9,81 m/s2).

Homework Equations

Calculate the work done when the box is drawn with a horizontal force.

The Attempt at a Solution

W = F*d
i don't know how to find F
i could only find F_f which is the frictional force
F_f= 0.58*182*9.81
F_f= 1035.54 N

 

[tiny-tim]

hi chawki!

W = F*d
i don't know how to find F
i could only find F_f which is the frictional force
F_f= 0.58*182*9.81
F_f= 1035.54 N

the speed is constant, so the acceleration is zero,

so F_total = 0,

so F + F_r = 0

 

[supratim1]

applied force is equal to force of friction in magnitude. so job is done!

 

[chawki]

yesyes,
by applying Newton second law and by projection on xx, we get:
F-F_f=0
F=F_f
F=0.58*182*9.81
F=1035.5436 N

W=F*d
W=1035.5436*21
W=21746.4156 J

 

[rwisz]

To calculate the work done, we need to find the net force acting on the box. Since the box is moving at a constant speed, we know that the net force is equal to zero. This means that the horizontal force applied to the box (F) must be equal in magnitude to the frictional force (Ff). Therefore, we can use the frictional force that you calculated (1035.54 N) as the horizontal force in the equation W = F*d.

W = 1035.54 N * 21 m = 21,722.34 Joules.

This is the work done by the horizontal force in moving the box along the floor. It is important to note that this work is only done against the frictional force, as there is no change in the box's kinetic energy (since it is moving at a constant speed).