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Answer :
To determine the stretch of the spring under a 10.1 kg board, we first calculate the force due to gravity which is 98.98 N. Using Hooke's law with the given spring constant of 176 N/m, the stretch of the spring is found to be 56.25 cm.
The question asks to find the amount by which a spring is stretched when supporting a 10.1 kg board that is parallel to the floor. The spring constant (k) is provided as 176 N/m.
Firstly, we need to find the force that the board exerts on the spring due to gravity. The force (F) due to gravity can be calculated using
F = m * g,
where 'm' is the mass of the board and 'g' is the acceleration due to gravity (which is approximately 9.8 m/s2).
For a mass of 10.1 kg, the force
F = 10.1 kg * 9.8 m/s2
= 98.98 N.
According to Hooke's law:
Fspring = -k * s
Where 'Fspring' is the force exerted by the spring, 'k' is the spring constant, and 's' is the displacement of the spring from its equilibrium position.
Using the given spring constant and the calculated force, we can rearrange the equation to solve for 's':
s = Fspring / k
s = 98.98 N / 176 N/m
= 0.5625 m or 56.25 cm.
Therefore, the spring is stretched 56.25 cm to support a 10.1 kg board.
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