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## Springs

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**Springs**Principles of Physics - Foederer**Elastic Potential Energy**Energy is stored in a spring when work is done to compress or elongate it Compression or elongation = change in length = x (m)**Springs have a spring constant (k) that tells how much force**is required to change the length of a spring by 1m. Robert Hooke put it all together→ Fs = kx Fs = spring force (N) k = spring constant (N/m) x = change in length (m)**Based on Hooke’s experiments, it was determined that the**energy stored in the spring can be calculated using the following formula: PEs = ½ kx2 PEs = elastic potential energy (joules) k = spring constant (N/m) x = change in length (m)**Example**A spring (k = 340 N/m) is compressed 0.02m. How much energy is stored in the spring? PEs = ½ kx2 = ½(340 N/m)(0.02m)2 = 0.068 J**Example**A 5 kg block sliding on a frictionless horizontal surface with a speed of 5 m/s collides with a spring (k = 120 N/m) and comes to rest. How much will the spring be compressed? KE→PEs ½ mv2 = ½ kx2 ½ (5kg)(5 m/s)2 = ½(120 N/m)(x)2 x = 1.02 m**Example**The diagram at right shows a 0.1-kilogram apple attached to a branch of a tree 2 meters above a spring on the ground below. The apple falls and hits the spring, compressing it 0.1 meter from its rest position. Assume all energy is conserved within the system. • Write the energy transformations that occur as the apple falls from the tree. ______→ ______→ ______ PE KE PEs**Example**The apple (0.1 kg) falls 2m and hits the spring, compressing it 0.1 meter from its rest position. Assume all energy is conserved within the system. • Determine the gravitational PE of the apple before it falls. PE= mgh PE = 0.1(9.8)(2) PE = 1.96 J**Example**The apple (0.1 kg) falls and hits the spring, compressing it 0.1 meter from its rest position. Assume all energy is conserved within the system. • Determine the kinetic energy and the velocity of the apple when it hits the spring. KE = ½ mv2 1.96 = ½ (0.1)(v2) ½ (0.1) ½ (0.1) 39.2 = v2 v = 6.26 m/s √ √**Example**The apple (0.1 kg) falls and hits the spring, compressing it 0.1 meter from its rest position. Assume all energy is conserved within the system. • Determine the elastic potential energy and the spring constant of this spring? PEs = ½ kx2 1.96 = ½(k)(0.1)2 ½ (0.1)2 ½ (0.1)2 k = 392 N/m