Suppose a force of 40 N is required to stretch and hold a spring 0.1 m from its equilibrium position. a. Assuming the spring obeys Hooke's law, find the spring constant k. b. How much work is req...
3) Consider Hooke's Law: The force required to keep a spring in a compressed or stretched position x units from the spring's equilibrium position is F(x)-kr Calculate the work required, in joules, to stretch a spring 0.4 meters beyond its equilibrium position for each of the following scenarios. a) The spring requires 50 Newtons of force to hold it 0.1 m from its equilibrium position. b) The spring requires 2 Joules of work to stretch the spring 0.1 meter from...
Consider a mass m suspended from a massless spring that obeys Hooke's Law (i.e. the force required to stretch or compress it is proportional to the distance stretched/compressed). The kinetic energy T of the system is mv2/2, where v is the velocity of the mass, and the potential energy V of the system is kr-/2, where k is the spring constant and x is the displacement of the mass from its gravitational equilibrium position. Using Lagrange's equations for mechanics (with...
6. A force of 30 N will stretch a spring 75cm (0.75m). Assuming Hooke's law applies, how far will a 110-N force stretch the spring? How much work does it take to stretch the spring this far?
To understand the use of Hooke's law for a spring. Hooke's law states that the restoring force F⃗ on a spring when it has been stretched or compressed is proportional to the displacement x⃗ of the spring from its equilibrium position. The equilibrium position is the position at which the spring is neither stretched nor compressed. Recall that F⃗ ∝x⃗ means that F⃗ is equal to a constant times x⃗ . For a spring, the proportionality constant is called the spring constant and denoted...
Caloulate the work required to stretch the following springs 1.5 m from their equilibrium positions. Assume Hooke's law is obeyed a. A spring that required 80 J of work to be stretched 0.2 m from its equilibrium position. b.A spring that required a force of 250 N to be stretched 0.4 m from its equibrium position a. Set up the integral that gives the work done in strotching the spring 1.5 m from its equilibrium position. Use increasing Imits of...
(1 point) Finding the work done in stretching or compressing a spring. Hooke's Law for Springs. According to Hooke's law, the force required to compress or stretch a spring from an equilibrium position is given by F(x) = kx, for some constant k. The value of k (measured in force units per unit length) depends on the physical characteristics of the spring. The constant k is called the spring constant and is always positive. Part 1. Suppose that it takes...
021 (part 1 of 2) 10.0 points The force required to stretch a Hooke's-law spring varies from 0 N to 69.3 N as we stretch the spring by moving one end 9.11 cm from its unstressed position. Find the force constant of the spring Answer in units of N/m. 022 (part 2 of 2) 10.0 points Find the work done in stretching the spring Answer in units of J.
Consider a spring that does not obey Hooke's law very faithfully. One end of the spring is fixed. To keep the spring stretched or compressed an amount x, a force along the x-axis with x-component Fx=kx−bx2+cx3 must be applied to the free end. Here k=100N/m, b=700N/m2, and c=12000N/m3. Note that x>0 when the spring is stretched and x<0 when it is compressed. A)How much work must be done to stretch this spring by 0.050 m from its unstretched length? B)How...
When a 3.80-kg object is hung vertically on a certain light spring that obeys Hooke's law, the spring stretches 2.30 cm. (a) If the 3.80-kg object is removed, how far will the spring stretch if a 1.50-kg block is hung on it? x When an object is 'hung' from a spring, the spring force is equal in magnitude to the gravitational force. cm (6) How much work must an external agent do to stretch the same spring 4.00 cm from...
Question 15 A spring required a force of 5.0 N to compress it 0.1 m. How much work is required to stretch the spring 0.4 m? Hint: Work done is equal to the change in elastic potential energy 0.8) O 0.4) 2) 0.2