A crate of mass 200 kg is pushed a distance of 20 m across a level floor. The crate is pushed with a force of 150N. The force of friction acting on the crate is 50N. The work done in pushing the crate across the floor is: A 1000J B 2000J C 3000J D 400 o J E 200 oo J

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A crate of mass 200 kg is pushed a distance of 20 m across a level floor. The crate is pushed with a force of 150N. The force of friction acting on the crate is 50N. The work done in pushing the crate across the floor is: A 1000J B 2000J C 3000J D 400 o J E 200 oo J

Answer 1

{\[Work = NetForce X Distance = 100N \times 20m = 2000 J\] Answer choices B: 2000 J is correct.}

Answer 2

## Step 1: Before we proceed with solving the posed problem, let's make a quick inventory of the important forces at play here: the worker's applied force and friction. Since the box is moved along a level floor, both gravity and the floor's normal force directly cancel each other out (acting perpendicular to the direction of motion), hence not contributing to any performed work. ### **Formula:** ### **Work = Force x Distance** ## Step 3: As within the example provided, we need to use the formula of work done i.e., "Force times Distance" in which both counterparts need to be parallel/have identical paths. But, in this problem the applied force (150 N) is acting exactly in the way we need i.e., we are only considering the horizontal component. So, the equation will be the product of the absolute value of the applied force (150 N), friction force (50 N) and the distance (20 m). ## Step 4: The applied force and the friction act in two opposite directions in our case, hence resulting in the net force being the subtraction of the values i.e., \((150\ \mathrm{N} - 50\ \mathrm{N}) = 100\ \mathrm{N}\). ## Step 5: Finally crucial observable made here is that there is no division of normal force or gravitational force in play while calculation for work, since both will completely negate each other due to their perpendicular relationship towards the crate's movement direction. Therefore, "work done by gravity" and the "work done by normal force" components will both yield **0 J (joules)** for us. ## Step 6: Calculate the final work value by multiplying the net force calculated in Step 4 and the distance short isn't by the crate.

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A crate of mass 200 kg is pushed a distance of 20 m across a level floor. The crate is pushed with a force of 150N. The force of friction acting on the crate is 50N. The work done in pushing the crate across the floor is: A 1000J B 2000J C 3000J D 400 o J E 200 oo J

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