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Vilka [71]
2 years ago
5

POSSIBLE POINTS: 60

Physics
1 answer:
Dima020 [189]2 years ago
4 0

Answer:

B. 57.4 N horizontal (right)

<u>Explanation</u>:

using cos method:

cos(35) = \frac{vertical force}{100}

vertical force = 57.4 N

<em>As its vertical, the direction of this force should be up.</em>

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A 540 gram object is attached to a vertical spring, causing the spring’s length to change from 70 cm to 110 cm.
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Answer:

Approximately 13\; {\rm N \cdot m^{-1}} (assuming that g = 9.81\; {\rm N \cdot kg^{-1}}.)

Explanation:

Let F_{\text{s}} denote the force that this spring exerts on the object. Let x denote the displacement of this spring from the equilibrium position.

By Hooke's Law, the spring constant k of this spring would ensure that F_\text{s} = -k\, x.

Note that the mass of the object attached to this spring is m = 540\; {\rm g} = 0.540\; {\rm kg}. Thus, the weight of this object would be m\, g = 0.540\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \approx 5.230\; {\rm N}.

Assuming that this object is not moving, the spring would need to exert an upward force of the same magnitude on the object. Thus, F_{\text{s}} = 5.230\; {\rm N}.

The spring in this question was stretched downward from its equilibrium by:

\begin{aligned} x &= (70\; {\rm cm} - 110\; {\rm cm}) \\ &= (-40)\; {\rm cm} \\ &= (-0.40) \; {\rm m}\end{aligned}.

(Note that x is negative since this displacement points downwards.)

Rearrange Hooke's Law to find k in terms of F_{\text{s}} and x:

\begin{aligned} k &= \frac{F_{\text{s}}}{-x} \\ &\approx \frac{5.230\; {\rm N}}{-(-0.40)\; {\rm m}} \\ &\approx 13\; {\rm N \cdot m^{-1}}\end{aligned}.

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