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barxatty [35]
2 years ago
13

A box is being dragged across the floor at a constant speed by a rope pulling horizontally on it. friction is not negligible. id

entify all the forces acting on the box.
Physics
1 answer:
Elena-2011 [213]2 years ago
8 0

Weight (W), the normal force( N), and the tension force( T )in the string will all be forces operating on the box.

<h3>What is tension force?</h3>

The tension force is described as the force transferred through a rope, string, or wire as opposing forces pull it.

The tension force is applied along the whole length of the wire, pulling energy equally on both ends.

A horizontally pulling rope is dragging a box across the floor at a consistent speed. Friction is significant.

Hence, the force acting on the box will be weight W,  Normal force N, and the tension force T, in the string.

To learn more about the tension force refer to the link;

brainly.com/question/2287912

#SPJ1

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Please do number 25! Explain how you got your answer with detail to get Brainliest! Thank you!
Ad libitum [116K]
John weighs 200 pounds.
In order to lift himself up to a higher place, he has to exert force of 200 lbs.

The stairs to the balcony are 20-ft high.
In order to lift himself to the balcony, John has to do
(20 ft) x (200 pounds)  =  4,000 foot-pounds of work.

If he does it in 6.2 seconds, his RATE of doing work is
(4,000 foot-pounds) / (6.2 seconds)  =  645.2 foot-pounds per second.

The rate of doing work is called "power".

(If we were working in the metric system (with SI units),
the force would be in "newtons", the distance would be in "meters",
1 newton-meter of work would be 1 "joule" of work, and
1 joule of work per second would be 1 "watt".
Too bad we're not working with metric units.)

So back to our problem.

John has to do 4,000 foot-pounds of work to lift himself up to the balcony,
and he's able to do it at the rate of 645.2 foot-pounds per second.

Well, 550 foot-pounds per second is called 1 "horsepower".

So as John runs up the steps to the balcony, he's doing the work
at the rate of

           (645.2 foot-pounds/second) / (550 ft-lbs/sec per HP)

=  1.173 Horsepower.  GO JOHN !

(I'll betcha he needs a shower after he does THAT 3 times.)
_______________________________________________

Oh my gosh !  Look at #26 !  There are the metric units I was talking about.

Do you need #26 ?

I'll give you the answers, but I won't go through the explanation,
because I'm doing all this for only 5 points.

a).  5
b).  750 Joules
c).  800 Joules
d).  93.75%

You're welcome.

And #27 is 0.667 m/s .
7 0
3 years ago
While moving a box 10.0 m across the floor, Sam pushes with a force of 150 N to the right. The force friction acting on the box
svet-max [94.6K]

Answer:

Explanation:required formula is

W 1=F*S

W1=work done by Sam =?

F=force applied by sam=150N

S=displacement =10m

again

W2=F*S

W2=work done by friction =?

S=displacement =10m

F=friction =25N

W=W1-W2=net work done

please feel free to ask if you have any questions

4 0
3 years ago
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