A 1.50 kg mass is placed on the end of a spring that has a spring constant of 17
ID: 1691254 • Letter: A
Question
A 1.50 kg mass is placed on the end of a spring that has a spring constant of 175 N/m. The mass-spring system rests on a frictionless incline that is at an angle of 30 degrees from the horizontal. The system is eased into its equilibrium position, where it stays. a) Determine the change in elastic potential energy of the system. b) Determine the system's change in gravitational potential energy. -In the first part, why do you use sin? instead of cos?, or add cos and sin together? Why is just mgsin?=kx for both part a and b? -And in part b, it is asking for the gravitational force. Why wouldn't you use PE=mgy?Explanation / Answer
Hey here the weight of the object mg is making an angle ? with the inclined surface. So it can be resolved into components. When we resolve the weight mg into it's components then mg cos ? balances with the normal reaction N . And mg sin? with the restoring forceof the spring . Means mgcos? = N and mgsin? = kx . This is how we will resolve any vector. PE = mgh is not the gravitational force but it is gravitational potential energy . And in the question he asked about the change in the gravitaional potential energy only . As the block is on a inclined plane instead of using mg here mgsin? is used and in place of h the distance the block moved x is used . So that the work done is nothing but the change in the potential energy . And is given by W = mgsin?(-x) -x is due to that spring expands and it's gravitational potential energy decreases.Related Questions
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