A mass of 100 g causes a vertical spring to stretch by 2.0 cm. a) find the spring constant of the spring in N/m: No picture was given to me. Ok im not sure if this is the formula to find spring constant so is it: k= mg/x and if so the answer is k=(100)(9.81)/2.0 cm k= 490.5 N/m is this correct?
The formula (Hooke's law) is correct. But check your units. I recommend that you convert all quantities to standard units before plugging numbers into the formula. (What are the standard units for length and mass?)
Ok here is the correct solution. k=mg/x k= (.1kg)(9.81)/.02m k= 49.1 N/m so I got the correct answer by using hooke's law? Now that I got that then for part b it asks: How much force wil cause the spring to compress by 0.5 cm? F=-kx So the answer is F= (-49.1N/m)(.005m)= -.25 N Did I do it right?
Good. Yes, except for that minus sign. Just give the magnitude of the force. (The meaning of the minus sign in Hooke's law is just that the force exerted by the spring is opposite to the displacement from equilibrium. If you pull the spring to the right, the spring force pulls back to the left. Don't get hung up on this.)
similar question: What is the spring constant of a spring that stores 25 J of elastic potential energy when compressed by 6.0 cm from its relaxed length? I've tried using N*m = J (work) to find my Newtons and dividing by the distance: N*0.06m = 25J N = 25 / 0.06 N = 416.667 N/m = 416.667N / 0.06m N/m = 6944.444 for some reason, that's wrong. I'm wondering if perhaps the sign matters?
Energy stored in a spring I assume you mean Work = Force x distance. That only works if the force is constant, but that's not true here. The more the spring is compressed, the stronger it pushes back. Either integrate ([itex]dW = F(x) dx[/itex]) or look up the formula for the energy stored in a compressed spring.
The force required to stretch a Hooke’s-law spring varies from 0 N to 52.4 N as we stretch the spring by moving one end 13.8 cm from its unstressed position. Find the force constant of the spring. Answer in units of N/m. Cant i just Take Force final = -k(change in x)
Hi, so what if the spring isn't vertical? I have the mass, displacement, and frequency of the spring and I'm being asked to find the acceleration. Do I need to find the spring constant in order to do that?
on what factors spring constant depends? for example if we attatch mass to a particular spring and calculate frequency, then we cut the spring into two halves and attatch the same mass to one of the halves. is there any difference in frequency or not then? please help me out.
so im doing a real life situation lab. but with poppers. (you know the little rubber ones you put inside out and let pop up and you go AH!). and all im asking is the x. when it's "Stretch like in the original problem asked in this thread would that be the hight compressed (i dont think so since it's 'compressed') the average height? Im guessing the average height
Hi I need help in one of these exam questions from the AQA Physics a level exam in springs constant.Unit 02 - Mechanics, Materials and Waves Question Paper on question 1aiii) http://www.aqa.org.uk/qualifications/a-level/science/physics-a/physics-a-key-materials PLEASE HELPP!!!
A teddy bear of mass 400 grams is hung from the end of a spring. The spring measures 51.0cm long in the rest position; when the teddy bear is attached to the end of the spring, the spring extends to 72.0cm. Calculate the elastic potential energy stored in the spring when extended to 72.0cm.