How much work did the movers do (horizontally) pushing a 46.0-kg crate 10.4 m across a rough floor without acceleration, if the effective coefficient of friction was 0.60? Express your answer using two significant figures.

The total work done by the mover is 2.81 kJ.

Explanation:

Given the 46 kg crate is displaced by 10.4 meters.

And the acceleration is zero. Also,

let is applied force, is the net force, is the mass and

and

As the acceleration is zero, the net force will also be zero.

.

Now, we know the work done is force times displacement.

So,

So, the total work done by the mover to displace 46 kg crate by 10.4 meters 2.81 kJ.

  distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

  slope of a graph = change in value of y -axis / change in values of x -axis

  comparing both the equations

    change in value of y -axis = change in position

  change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.