You use the equation F=1/2(kx^2). When you plug your numbers in you should get
F=(1/2)(8.25N/m)(0.150m)^2= 0.0928
3. A spring has a spring constant k=8.75 N/m. If the spring is displaced 0.150 m from its equilibrium position, what i...
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 required to compress the spring 0.2 m from its equilibrium position? c. How much work is required to stretch the spring 0.5 m from its equilibrium position? d. How much additional work is required to stretch the spring 0.1 m if it has...
A spring of force constant 3 N/m is compressed by 5.0 em from its equilibrium position. The spring is then released and stretched by 5.0 ㎝ from its equilibrium position. Find the difference in potential energy between the two positions of the spring (stretched and compressed).
A spring is suspended vertically from a fixed support. The spring has spring constant k=24 N m −1 k=24 N m−1 . An object of mass m= 1 4 kg m=14 kg is attached to the bottom of the spring. The subject is subject to damping with damping constant β N m −1 s β N m−1 s . Let y(t) y(t) be the displacement in metres at the end of the spring below its equilibrium position, at time t...
A light spring with a spring constant k = 316 N/m is attached to a vertical wall at one end and a block with a mass m = 0.462 kg at the other end. The block rests on a horizontal frictionless surface and is initially at the equilibrium length of the spring. The block is then displaced from the equilibrium position of the spring in such a manner as to stretch the spring by an amount A = 0.190 m...
A spring is suspended vertically from a fixed support. The spring has spring constant k=24 N m −1 k=24 N m−1 . An object of mass m= 1 4 kg m=14 kg is attached to the bottom of the spring. The subject is subject to damping with damping constant β N m −1 s β N m−1 s . Let y(t) y(t) be the displacement in metres at the end of the spring below its equilibrium position, at time t...
SPRING +0000000 x=0 +AX 6.5 When a spring is stretched or compressed from its equilibrium position by an amount Ar it exerts a force given by F =-kar. The - sign indicates that this force is in the opposite direction of Ar and is oriented to restore the spring to its e k is a proportionality constant called the spring constant or the force constant and is related to the stiffness of the spring. A. Draw a force vs displacement...
5 points) A spring is suspended vertically from a fixed support. The spring has spring constant k=28 N m−1k=28 N m−1. An object of mass m=14 kgm=14 kg is attached to the bottom of the spring. The subject is subject to damping with damping constant β N m−1 sβ N m−1 s. Let y(t)y(t) be the displacement in metres at the end of the spring below its equilibrium position, at time tt seconds. (5 poins) A spring is suspended vertically...
To stretch a certain spring by 3.00 cm from its equilibrium position requires 7.50 J of work. Part A What is the force constant of this spring? k k = nothing N/m SubmitRequest Answer Part B What was the maximum force required to stretch it by that distance? F F = nothing N SubmitRequest Answer Return to AssignmentProvide Feedback
A 0.334-kg mass is attached to a spring with a force constant of 54.9 N/m Part A If the mass is displaced 0.200 m from equilibrium and released, what is its speed when it is 0.131 m from equilibrium? ΑΣφ m/s Previous Answers Request Answer Submit X Incorrect; Try Again; 5 attempts remaining Return to Assign ment Provide Feedback
A horizontal spring attached to a wall has a force constant of k = 900 N/m. A block of mass m = 1.30 kg is attached to the spring and rests on a frictionless, horizontal surface as in the figure below. (a) The block is pulled to a position xi = 5.20 cm from equilibrium and released. Find the potential energy stored in the spring when the block is 5.20 cm from equilibrium. 1.22J : Your answer is correct. (b)...