Question

please fill out the table with explanation

EXPERIMENT #12 HOOKE'S LAW AND SPRING DEFORMATION EXPERIMENT A measurement of mass(kg) attached to an extensible spring were conducted and recorded. Initially a mass in kg was attached to the spring, and the corresponding extension was measure with a meter stick recorded. Subsequent masses were added to mass already hanging on the spring and the table below was computed to give stress (Pa), and strain of the experiment. DATA FROM THE EXPERIMENT COPPER WIRE AND MASS Stress o (Pa) Strain E 0.0098 0.004 0.0294 0.011 0.0686 0.017 0.1176 0.029 0.147 0.032 0.2058 0.038 0.2646 0.0468 0.3038 0.0586 0.33516 0.0734 0.29008 0.0819 0.26558 0.092 0.28518 0.11 0.32732 0.123 0.34349 0.13 0.37926 0.15 0.42336 0.17 0.48216 0.2 0.47432 0.22 0.4312 0.233 0.40866 0.245 0.36554 0.26 Instruction: Plot the above data point in excel and answer the corresponding questions that follows. First Plot: Plot Stress vs. Strain Second Plot: Plot Stress vs. Strain for the first seven data points that shows direct proportionality as defined by Hooke's Law.

Answer 1

LINEAR THERMAL EXPANSION

$dL=\alpha L_odT$

to calculate the change in length, width and height (which are all one diemensional), we use linear thermal expansion.

VOLUME THERMAL EXPANSION m

$dv=\beta V_odT$

where $\beta$ is the coeffecient of volume thermal expansion and $\beta = 3*\alpha$

to calculate the change in volume(which is three diemensional), we use volume thermal expansion.

=>change in temperature, dT = T2 -T1 = (95-20) oC = 75 oC

material	coeffecient of linear thermal expansion, $\alpha$ ( per oC *10-6)	length, Lo(m)	width, Wo(m)	height, Ho(m)	dL (m)	dW(m)	dH(m)	dV(m3)
gold	14	0.35	0.07	0.07	367.5 * 10-6	73.5 * 10-6	73.5 * 10-6	5.4014 * 10-6
silver	18	0.25	0.05	0.05	337.5 * 10-6	67.5 * 10-6	67.5 * 10-6	2.53* 10-6
aluminium	24	0.40	0.08	0.08	720 * 10-6	144* 10-6	144* 10-6	13.824* 10-6

case 1: gold

$dL=\alpha L_odT$

dL = 14 * 10-6 * 0.35 * 75 = 367.5 * 10-6 m

dW = 14 * 10-6 * 0.07 * 75 = 73.5 * 10-6 m

dH = 14 * 10-6 * 0.07 * 75 = 73.5 * 10-6 m

$dv=\beta V_odT=3*\alpha V_odT=3*\alpha *(Lo*Wo*Ho)*dT$

dV = 3 * 14 * 10-6 * (0.35*0.07*0.07) * 75 = 5.4014 * 10-6 m3

case 2: silver

$dL=\alpha L_odT$

dL = 18 * 10-6 * 0.25 * 75 = 337.5 * 10-6 m

dW = 18 * 10-6 * 0.05 * 75 = 67.5 * 10-6 m

dH = 18 * 10-6 * 0.05 * 75 = 67.5 * 10-6 m

$dv=\beta V_odT=3*\alpha V_odT=3*\alpha *(Lo*Wo*Ho)*dT$

dV = 3 * 18 * 10-6 * (0.25*0.05*0.05) * 75 = 2.53* 10-6 m3

case 3: aluminium

$dL=\alpha L_odT$

dL = 24 * 10-6 * 0.4 * 75 = 720 * 10-6 m

dW = 24 * 10-6 * 0.08 * 75 = 144 * 10-6 m

dH = 24 * 10-6 * 0.08 * 75 = 144 * 10-6 m

$dv=\beta V_odT=3*\alpha V_odT=3*\alpha *(Lo*Wo*Ho)*dT$

dV = 3 * 24 * 10-6 * (0.40*0.08*0.08) * 75 = 13.824* 10-6 m3