Question

A load of 5kg stretches a coil spring to a length of 24cm and a load of 8kg stretches to a length of 30cm. find the length of the spring when there is no load

Answer 1

F= k x
F=mg

mg = kx

Answer 2

are assuming this spring is hanging or on a table?

Answer 3

physics question but i guess we can help..

Answer 4

hook's law is introduced in math as well as physics.

Answer 5

let "s" be the initial length of the spring x is the total displacement of the spring which means
x1 = .24-s
and
x2 = .30-s

Answer 6

I don't quite understand but thanks anyways

Answer 7

5kg stretches a coil spring to a length of 24cm

F= mg - equation for weight/force of an object where "m" is mass (kg) and g is the gravitational constant 9.81 m/s^2

F=kx - equation for force of a spring where "k" is the stiffness or spring constant and x is the total displacement of the spring or how far the spring is stretched

Answer 8

the force of the weight is equal to the force of the spring thus

mg = kx

Answer 9

now applying the first statement
5kg stretches a coil spring to a length of 24cm

and substituting hte values in to the equation
mg =kx

now, from above,
x = 24-s

substituting all values
5 (9.81) = k (.24 - s)

Answer 10

now do the same for the 2nd statement

Answer 11

once you have 2 equations, set both equations equal to k

then combine the 2 equations into 1 equation assuming that the stiffness or spring constant of the springs are the same

then solve for s which is the initial length of the spring

Answer 12

ok thanks so much that makes a lot more sense

Answer 13

no prob