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A 2000-kg ore car rolls 50 meters down a smooth 10o incline. There is a horizontal spring at the end of the incline designed to stop the car in case of break failure. What is the spring constant of the spring if it would compress by 1 m to stop the ore car?

Answer :

ONYEBUCHINNAJIIN

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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