Answer:
The value is [tex]k = 29.28 \ N / m[/tex]
Explanation:
From the question we are told that
The mass of the cart is m = 0.68 kg
The angle of inclination is [tex]\theta = 61.5^o[/tex]
The length of the spring when the cart is a equilibrium is [tex]d = 52.0 \ cm = 0.52 \ m[/tex]
The length of the spring when at rest is [tex]d_r = 12 \ cm = 0.12 \ m[/tex]
Generally the weight of the cart is mathematically represented as
[tex]W_c = m * g * sin (\theta )[/tex]
=> [tex]W_c = 0.68 * 9.8 * sin (61.5 )[/tex]
=> [tex]W_c = 5.856 \ N[/tex]
Generally the force applied on the spring by the weight of the cart is mathematically represented as
[tex]F = \frac{1}{2} * k * (d - d_r )[/tex]
So at equilibrium
[tex]W_c = F[/tex]
=> [tex]\frac{1}{2} * k * (d - d_r ) = 5.856[/tex]
=> [tex]0.5 * k * (0.52 - 0.12 ) = 5.856[/tex]
=> [tex]k = 29.28 \ N / m[/tex]
