Answered

Welcome to Westonci.ca, where finding answers to your questions is made simple by our community of experts. Get expert answers to your questions quickly and accurately from our dedicated community of professionals. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.

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?

Sagot :

onyebuchinnaji

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

#SPJ4

We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. We're dedicated to helping you find the answers you need at Westonci.ca. Don't hesitate to return for more.