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Nadusha1986 [10]
3 years ago
6

For a certain spring, k = 15N/m. A weight is hung from the spring, stretching it from 0.3m to 0.4m. What force did the weight pr

oduce
HELP PLS
Physics
1 answer:
Mashcka [7]3 years ago
8 0
The stretch of the spring is
\Delta x = 0.4 m-0.3 m=0.1 m
The constant of the spring is k=15 N/m, so we can find the force produced by the weight by using Hook's law:
F=k\Delta x=(15 N/m)(0.1 m)=1.5 N
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The system's tension is 616 N and acceleration is 5.6 m / s^{2}

<u>Explanation:</u>

From newton’s second law of motion which state that net force acting on a body is product of mass of a body and acceleration of a body which is given as,

             F_{n e t}=m_{t o t} \times a

Where,

F_{n e t} is net force acting on body

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Given values  

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                a=\frac{F_{n e t}}{m_{\mathrm{tot}}}=\frac{m \times g}{m_{\mathrm{tot}}}

Put the value for m = hanging mass = 40 kg and g=9.8 \mathrm{m} / \mathrm{s}^{2}, we get

                  a=\frac{40 \times 9.8}{30+40}=\frac{392}{70}=5.6 \mathrm{m} / \mathrm{s}^{2}

The tension in the ropes,  T=(m \times g)+(m \times a)

Here, m as hanging mass

T = tension, N or  k g m / s^{2}

m = mass, kg  

g = gravitational force, 9.8 \mathrm{m} / \mathrm{s}^{2}

a = acceleration, m / s^{2}

          T = (40 \times 9.8)+(40 \times 5.6) = 392+224 = 616 N

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