A 72.0-kg person pushes on a small doorknob with a force of 5.00 N perpendicular to the surface of the door. The doorknob is loc

The angle formed with the positive x-axis is 120 degrees. We can assume that this angle is determined in a counterclockwise direction from the positive x-axis. The x-component of the vector can be calculated as: x-component = 10 cos(120) = -5. The vector's y-component is determined as: y-component = 10 sin(120) = 8.66. The x-component equates to -5 while the y-component equals 8.66.

We know that F=ma
where m represents mass and a indicates acceleration
thus, Force= ma
therefore, F=1300X1.07=1391N
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Response:

The work performed by the particle traveling from x = 0 to x = 2 m totals 20 J.

Details:

The force impacting a particle, which is restricted to the x-axis, is expressed as follows:

We need to calculate the work done on a particle moving from x = 0.00 m to x = 2.00 m.

The formula for the work done by the particle is defined as:

Consequently, the work executed by the particle between x = 0 and x = 2 m amounts to 20 J. Thus, this is the solution sought.