Conveyor belts are often used to move packages around warehouses. The conveyor shown below moves packages at a steady 4.0 m/s. A 500 N employee decides to catch a ride by sitting on a 1,000 N box for a 50 m trip as shown below:



What is the work done by the employee on the box?

Answers

Answer 1

Answer:

0 j

Explanation:

Answer 2

The work done by the employee on the box at the given zero displacement is 0 J.

The given parameters;

Constant velocity of the conveyor, v = 4 m/sWeight of the employee, W = 500 NWeight of the box, W = 1,000 NDistance of the trip, h = 50 m

The work done by the employee on the box is calculated as follows;

W = Fd

where;

F is the applied force on the box by employee = weight of the employeed is the distance through which the box is moved

Since the employee sits on the box without moving it, the distance moved by the box = 0

W = 500 x 0

W = 0 J

Thus, the work done by the employee on the box is 0 J.

Learn more about work done and displacement here: https://brainly.com/question/8635561


Related Questions

Energy stored because of an object's height above the Earth's surface is_____energy.

nuclear

gravitational

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Answers

gravitational potential energy

A water-skier of mass 75.0 kg initially at rest is being pulled due east by a horizontal towrope. The rope exerts a force of 365 N (east). The water (and air) exerts a combined average frictional force of 190 N (in the opposite direction). How fast will the skier be moving after a distance of 38.0 m?

Answers

Answer:

The skier will be moving at 13.31 m/s.

Explanation:

To calculate the velocity of the skier we need to find the acceleration, as follows:

[tex] \Sigma F = ma [/tex]

[tex] F_{r} - F_{f} = ma [/tex]

Where:

[tex] F_{r}[/tex]: is the force due to the rope = 365 N

[tex] F_{f}[/tex]: is the combined average frictional force = 190 N

m: is the mass = 75.0 kg

[tex] a = \frac{365 N - 190 N}{75.0 kg} = 2.33 m/s^{2} [/tex]

Now, we can calculate the velocity of the skier by using the following kinematic equation:

[tex] v_{f}^{2} = v_{0}^{2} + 2ad [/tex]

Where:

[tex] v_{f}[/tex]: is the final velocity =?

[tex] v_{0}[/tex]: is the initial velocity = 0 (the skier is initially at rest)

d: is the distance = 38.0 m

[tex] v_{f} = \sqrt{2*2.33 m/s^{2}*38.0 m} = 13.31 m/s [/tex]

Therefore, the skier will be moving at 13.31 m/s.

I hope it helps you!                                                                                  

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