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VLD [36.1K]
3 years ago
6

A load of 54 N attached to a spring that ishanging vertically stretches the spring 0.15 m.What is the spring constant?Answer in

units of N/m.
Physics
2 answers:
beks73 [17]3 years ago
8 0

Answer:

300 N/m

Explanation:

given,

Load attached to the spring, W = 54 N

length of stretch of the spring, x = 0.15 m

spring constant= ?

Force applied on the spring is calculated by the equation

F = k x

where k is the spring constant

x is the displacement of the spring due to applied load

now,

54 = k × 0.15

k = \dfrac{54}{0.15}

k =300\ N/m

hence, the spring constant is equal to 300 N/m

myrzilka [38]3 years ago
6 0

Answer: The spring constant is 360 N/m

Explanation:

To calculate the spring constant, we use the equation:

F=kx       ......(1)

where,

F = force exerted on the spring = 54 N

k = spring constant = ?

x = length of the spring = 0.15 m

Putting values in equation 1, we get:

54N=k\times 0.15m\\\\k=\frac{54N}{0.15m}=360N/m

The spring constant is 360 N/m

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1. Two forces act on a box as follows: F1 = 100 N at 01 = 170° and F2 = 75 N
lilavasa [31]

Answer:

a)  F = 64.30 N,  b) θ = 121.4º

Explanation:

Forces are vector quantities so one of the best methods to add them is to decompose each force and add the components

let's use trigonometry

Force F1

          sin 170 = F_{1y} / F₁

          cos 170 = F₁ₓ / F₁

          F_{1y} = F₁ sin 170

          F₁ₓ = F₁ cos 170

          F_{1y} = 100 sin 170 = 17.36 N

          F₁ₓ = 100 cos 170 = -98.48 N

Force F2

          sin 30 = F_{2y} / F₂

          cos 30 = F₂ₓ / F₂

          F_{2y} = F₂ sin 30

          F₂ₓ = F₂ cos 30

          F_{2y} = 75 sin 30 = 37.5 N

          F₂ₓ = 75 cos 30 = 64.95 N

the resultant force is

X axis

          Fₓ = F₁ₓ + F₂ₓ

          Fₓ = -98.48 +64.95

          Fₓ = -33.53 N

Y axis

         F_y = F_{1y} + F_{2y}

         F_y = 17.36 + 37.5

         F_y = 54.86 N

a) the magnitude of the resultant vector

let's use Pythagoras' theorem

         F = Ra Fx ^ 2 + Fy²

         F = Ra 33.53² + 54.86²

         F = 64.30 N

b) the direction of the resultant

let's use trigonometry

        tan θ’= F_y / Fₓ

        θ'= tan^{-1}  \frac{F_y}{F_x}

        θ'= tan⁻¹ (54.86 / (33.53)

        θ’= 58.6º

this angle is in the second quadrant

The angle measured from the positive side of the x-axis is

        θ = 180 -θ'

        θ = 180- 58.6

        θ = 121.4º

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A 19kg block is being pulled with a constant horizontal force of 95 Newton’s while also experiencing a constant friction force o
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Answer:

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Explanation:

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