Aspring on a horizontal surface can be stretched and held 0.6 m from its equilibrium position with a force of 78 n. a. how much work is done in stretching the spring 5.5 m from its equilibrium position? b. how much work is done in compressing the spring 1.5 m from its equilibrium position?

Step-by-step explanation:

x = 0.6 m

F = 78 N

Let k be the spring constant.

F = k x

78 = 0.6 k

k = 130 N/m

(a)

x = 5.5 m

W = 0.5 kx²

W = 0.5 x 130 x 5.5 x 5.5

W = 7865 J

(B) x = 1.5 m

W = 0.5 kx²

W = 0.5 x 130 x 1.5 x 1.5

W = 146.25 J

i think it is 70

step-by-step explanation:

x =±sqrt(1/3)

step-by-step explanation:

x+4/6x=1/x

multiply each side by 6x to get rid of the fractions

6x*(x+4/6x)=1/x*6x

distribute

6x^2 +4 = 6

subtract 4 from each side

6x^2 +4-4 = 6-4

6x^2 =2

divide by 6

6x^2/6 = 2/6

x^2 = 1/3

take the square root of each side

sqrt(x^2) = ±sqrt(1/3)

x =±sqrt(1/3)