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Sagot :
The spring constant of the spring if it would compress by 1 m to stop the ore car is 340,350.4 N/m.
Potential energy of the car
The potential energy of the car is calculated as follows;
P.E = mgh
where;
- h is the vertical height of the incline
sin (10) = h/L
h = L x sin(10)
P.E = mgLsin(10)
P.E = 2000 x 9.8 x 50 x sin(10)
P.E = 170,175.21
Conservation of energy
The potential energy of the car at the top of the incline will be converted to the elastic potential energy spring at the bottom of the incline.
P.E = U = ¹/₂kx²
k = (2U)/x²
k = (2 x 170,175.2)/(1)²
k = 340,350.4 N/m
Learn more about spring constant here: https://brainly.com/question/1968517
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