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Mariulka [41]
3 years ago
10

II. What examples can you find in your home that are examples of kinetic and potential energy (name two for each type

Physics
1 answer:
Trava [24]3 years ago
7 0

Answer:

11. A door accelerating when pushing it.

12. A person sliding along the floor.

13. Water dropping from the faucet.

14. Courtains descending when loosening them.

Explanation:

For kinetic energy, any example that does not move horizontally will do. Something accelerating or decelerating is a good example (Ek=1/2mv^2, where v is velocity).

For potential energy try to think of things moving vertically. Potential energy comes from the height of an object (Ep=mgh, where h is height).

Notice that example 13 and 14 represent potential energy turning into kinetic energy, height becomes lower, so velocity increases.

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A worker pushes horizontally on a 35.0 kg crate with a force of magnitude 112 N. The coefficient of static friction between the
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Explanation:

a) Friction force always oppose to the relative movement between two surfaces, and, provided that be less than the fs max, adopt any value to counteract the applied force.

The fs max, is the horizontal component of the contact force, and can be written as follows:

Fs max = us . Fn  

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b) Now, as the applied force is smaller than Fs max, this means that the friction force, is equal and opposite to the applied forcé, i.e., -112 N, so the crate doesn´t move.

c) Please see above.

d) As explained above, the maximum friction force, is proportional to the normal force, which adopts any value needed to satisfy the Newton´s 2nd Law.

So, if we diminish the normal force, we can lower the máximum friction force, helping to the worker to move the crate.

The mínimum needed normal force, will be the one that satisfies the following:

Fs max = F applied = us Fn = 112 N = 0.37. Fn.

Solving for Fn, we get Fn= 303 N

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Fup = 343 N -303 N = 40 N  

e) If we keep the normal force unchanged, but add an horizontal force to help the worker, we will need that the sum of both forces, will be equal to the Fs max, as follows:

Fh + Fapp = 127 N → Fh = 127 N – 112 N = 15 N

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