by using the principle of condervation of energy i had to find the spring consta
ID: 1686821 • Letter: B
Question
by using the principle of condervation of energy i had to find the spring constant of the cord. then they aske me to look for the location of the equilibrium point where the spring force balances the gravitational force exerted on the jumper. (this point is taken as the origin in our mathematical description of simple harmonic oscillation. when i went to see the sollition of this question, cramter started saying that at equilibrium position xk=mg?... :( I just took physics I three years ago.. i dont remember. Given data: bunge jumper: 65 kg. unstrtched length of the cord: 11 m the jumper reaches the bottom of her motion 36m below the bridge before bouncing back. thanks so mcuh! by using the principle of condervation of energy i had to find the spring constant of the cord. then they aske me to look for the location of the equilibrium point where the spring force balances the gravitational force exerted on the jumper. (this point is taken as the origin in our mathematical description of simple harmonic oscillation. when i went to see the sollition of this question, cramter started saying that at equilibrium position xk=mg?... :( I just took physics I three years ago.. i dont remember. Given data: bunge jumper: 65 kg. unstrtched length of the cord: 11 m the jumper reaches the bottom of her motion 36m below the bridge before bouncing back. thanks so mcuh!Explanation / Answer
When the jumper is at a distance less than the unstretched length of the string, there is no effect of the weight applied on the string. i.e. tension in the string is zero upto that level. When the jumper just crosses the length of the unstretched string then the tension in the string developed against the free fall of the jumper. Thus the tension in the string will be compensated exactly by the weight of the jumper at the equilibrium position. Let x be the elongation observed in the string at the equilibrium, then the tension in the string is kx (k is the force constant) which is equal to the weight mg of the jumper Thus xk=mg
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.