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jekas [21]
3 years ago
11

A box that has a weight of 40 N is hung from a spring. The spring constant of the spring is 400 N/m.

Physics
2 answers:
USPshnik [31]3 years ago
6 0

Answer:

the spring will stretch 10 cm

Explanation:

With the 40 N mass hanging from it, the restorative force of the spring will equal 40 N.

Let's recall that the restorative force "F" of the spring has a magnitude proportional to the displacement "x" from its relax position as per the following formula:

F = k* x where "k" is the spring's constant.

We can from this expression find the value of the displacement "x" just solving for it in the equation:

F = k* x\\40\,N=400 \frac{N}{m} *x\\x=\frac{40}{400} m\\x=0.1\,m

Such displacement is equivalent to 10 cm.

Elena L [17]3 years ago
4 0

Answer:

The correct answer is 10 cm.

Explanation:

Just did the quiz on edg. :)

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4 years ago
A box is being dragged with a horizontal force of 65 N for 12 meters. If there is a force of friction acting on it
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Answer:

A. 780 J

B. 120 J

C. 660 J

Explanation:

From the question given above the following data were obtained:

Dragging force (Fₔ) = 65 N

Distance (s) = 12 m

Force of friction (Fբ) = 10 N

A. Determination of the work done by the dragging force.

Dragging force (Fₔ) = 65 N

Distance (s) = 12 m

Workdone (Wd) by dragging force =?

Wd = Fₔ × s

Wd = 65 × 12

Wd = 780 J

Therefore, the work done by the dragging force is 780 J

B. Determination of the work done by friction.

Distance (s) = 12 m

Force of friction (Fբ) = 10 N

Workdone (Wd) by friction =?

Wd = Fբ × s

Wd = 10 × 12

Wd = 120 J

Therefore, the work done by friction is 120 J

C. Determination of the net work done on the box.

Dragging force (Fₔ) = 65 N

Distance (s) = 12 m

Force of friction (Fբ) = 10 N

Net work done (Wd) =?

Next, we shall determine the net force acting on the box. This can be obtained as follow:

Dragging force (Fₔ) = 65 N

Force of friction (Fբ) = 10 N

Net force (Fₙ) =?

Fₙ = Fₔ – Fբ

Fₙ = 65 – 10

Fₙ = 55 N

Thus, the net force acting on the box is 55 N

Finally, we shall determine the net work done on the box as follow:

Distance (s) = 12 m

Net force (Fₙ) = 55 N

Net work done (Wd) =?

Wd = Fₙ × s

Wd = 55 × 12

Wd = 660 J

Therefore, the net work done on the box is 660 J

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