Transcript Document
Spring Potential Energy When a mass is hung from the end of a spring, the spring stretches an amount x from the resting position. The force of gravity acting on the object does work on the spring to stretch it. After a mass is hung, when the system is at rest, it is at equilibrium. This means the downward force of gravity is balanced by the upward restoring force of the spring. It should make sense that a large force is required to compress or expand a stiff spring a certain distance while less force is needed to compress or expand a “loose” spring that same distance. An equation that relates the restoring force of the spring to the amount is stretches is summarized as Hooke’s Law: Fs = -kx Hooke’s Law explains how the distance a spring stretches relates to the force acting on the spring. Fs = -kx Fs is the restoring force, x is the displacement (distance) in meters k is the spring constant in N/m The negative on the right side of the equation simply means that the restoring force is in the opposite direction from the displacement. When the spring constant is a large number, it takes more force to stretch the spring a certain distance and we would consider the spring to be somewhat stiff. When the spring constant is a small number, it takes less force to stretch the spring a certain distance and we would consider the spring to be somewhat loose. As long as a spring is not stretched too far, the spring constant for a particular spring does not change and can be used to predict the displacement of the spring if we know the force applied. Question 11 If a 200 g mass displaces the spring 10 cm how much would the spring be displaced if a 400 g mass was hung from the spring? A simple proportion tells us the answer is ????? 200 g = 400 g 20 cm 10 cm ? Question 12 What is the spring constant of a spring if a force of 6 Newtons displaces it 0.3 meters Fs = - kx rearrange equation to solve for k so k = - Fs x k = 6 N = - 20 N/m 0.3m Question 13 What is the restoring force of a spring if it has a spring constant of 25 N/m and is stretched 40 cm from it’s original position? Fs = - kx Fs = - (25 N/m) (0.4 m) = - 10 N The negative sign simply means the force is directed upwards or opposite the force of gravity. When work is done to displace a spring a given distance x, the amount of potential energy stored in the spring can be calculated from the following equation: EPE = ½ kx2 Question 15 A stiff spring has a spring constant of 40 N/m. If the spring is displaced 0.15 m from rest what is the elastic potential energy of the spring? EPE = ½ kx2 EPE = ½ (40 N/m) (0.15 m)2 EPE = 0.45 Joules Question 17 An archer pulls a string on a bow a distance of 0.7 m from rest with a force of 140 N. What is the spring constant? Fs = - kx rearrange equation to solve for k so k = Fs x k = 140 N/m k = 200 N/m 0.7 m Question 18 What is the elastic potential energy stored in the bow? EPE = ½ kx2 EPE = ½ (200 N/m) (0.7 m)2 Question 19 How much kinetic energy is transferred to the arrow at the moment it leaves the bow? EPE = 49 J 49 J Question 20 If the arrow has a mass of .025 kg, what is the velocity of the arrow when it leaves the bow? KE = ½ mv2 V2 = 2 (49 J) .025 kg V = 62.6 m/s so v2 = 2(KE) m V2 = 3920