A0.15 kg mass is suspended from a vertical spring and descends a distance of 4.6 cm, after which it hangs at rest. an additional 0.50 kg mass is then suspended from the first. what is the total extension of the spring? (neglect the mass of the spring.)a. follow up: how much work is done by gravity in stretching the spring through both displacements?

The total extension = 19.9 cm

Work done in stretching the spring through both displacement = 0.633 J.

Explanation:

From Hook's law,

F  = ke Equation 1.

Where F = force or weight, k = force constant of the spring, e = extension.

making k the subject of the equation,

k = F/e Equation 2

Given: F = W = mg = 0.15(9.8) = 1.47 N, e = 4.6 cm = 0.046 m.

k = 1.47/0.046

k = 31.96 N/m.

When an additional mass of 0.5 kg is suspended,

Total mass suspended(M₁) = 0.15+0.5 = 0.65 kg

Total Weight (W₁) = 0.65(9.8) = 6.37 N.

k = 31.96 N/m

Substituting into equation 1

6.37 = 31.96e

e = 6.37/31.96

e = 0.199 m

e = 19.9 cm

Thus the total extension = 19.9 cm

a.

Work done in stretching the spring  through both displacement

W = 1/2ke² Equation 3

Where W = work done, k = spring constant, e = extension.

Given: k = 31.96 N/m, e = 0.199 m

Substituting into equation 3

W = 1/2(31.96)(0.199)²

W = 0.633 J.

two charges are fixed in place on the x-axis as shown. the charge at x = 0 is negative and has magnitude q = 2 μcoulombs (1 μcoulomb = 10-6 coulombs). the charge at x = d, (d = +12 cm), is positive and has magnitude q = 4 μcoulombs. find a point x = xo along the x-axis at which the electric field is zero.

answer is slate

hope this