Hooke's Law.docx

Hooke's Law In the diagram below is shown a block attached to a spring. In position (A) the spring is at rest and no external force acts on the block. In position (B) a force F is used to compress the spring by a length equal to Δ x by pushing the block to the left. In position (C), a force F is used to stretch the spring by a length Δ x by pulling the block to the right. Δ x is the change in the length of the spring measured from its position at rest. spring and Hooke's law

In both cases, the relationship between the magnitude of force F used to stretch or compress the spring by a length Δ x is given by Hooke's law as follows: | F | = k | Δx |

where k is the spring constant. According to Newton's third law, if a spring is stretched or compressed using force F, as a reaction the spring also react by a force - F. Problems with Detailed Solutions Problem 1 What is the magnitude of the force required to stretch a 20 cm-long spring, with a spring constant of 100 N/m, to a length of 21 cm? Solution The spring changes from a length of 20 cm to 21 cm, hence it stretches by 1 cm or | Δx | = 1 cm = 0.01 m. | F | = k | Δx | = 100 N / m × 0.01 m = 1 N Problem 2 What is the spring constant of a spring that needs a force of 3 N to be compressed from 40 cm to 35 cm? Solution The spring changes from a length of 40 cm to 35 cm, hence it stretches by 40 cm - 35 cm = 5 cm or | Δx | = 5 cm = 0.05 m. | F | = k | Δx | = 3 N k = | F | / |Δx| = 3 / 0.05 = 60 N / m Springs in Parallel

Two springs are said to be in parallel when used as in the figure below. spring in parallel and Hooke's law

The two springs behave like on spring whose constant k is given by k = k1 + k2 Problem 3 What is the magnitude of the force required to stretch two springs of constants k1 = 100 N / m and k2 = 200 N / m by 6 cm if they are in parallel? Solution The two springs behave like a spring with constant k given by k = k1 + k2 = 100 N / m + 300 N / m | F | = k | Δx | = 300 N / m × 0.06 m = 18 N Spring in Series Two springs are said to be in series when used as in the figure below. spring in parallel and Hooke's law

The two springs behave like on spring whose constant k is given by 1 / k = 1 / k1 + 1 / k2 Problem 4 What is the magnitude of the force required to stretch two springs of constants 100 N / m and 200 N / m by 6 cm if they are in series? Solution The two springs behave like a spring with constant k obtained by solving for k the following 1 / k = 1 / 100 + 1 / 300 k = 75 N / m | F | = k | Δx | = 75 N / m × 0.06 m = 4.5 N Potential Energy of a Spring To stretch or compress a spring by a length | Δx |, energy is needed. Once stretched or compressed, the energy is stored in the spring as potential energy Pe and is given by: Pe = (1/2) k (Δx)2 , k is the spring constant. M

Problem 5 How much energy W is needed to compress a spring from 15 cm to 10 cm if the constant of the spring is 150 N / m? Solution Δx = 10 cm - 15 cm = - 5 cm = - 0.05 m The energy W to compress the spring will all be stored as potential energy Pe in the spring, hence W = Pe = (1/2) k (Δx)2 = 0.5 × 150 × (- 0.05)2 = 0.1875 J Custom Search search Share

Popular Pages Home