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//Given that
m1 = 0.70 //in kg
m = [0.14, 3.2] //in kg
k = [4.1* 10^4, 2.6* 10^6] //in N/m
d = [16* 10^-3, 1.1* 10^-3] //in meter
//Sample Problem 10-3a
printf("**Sample Problem 10-3a**\n")
name = ['board', 'block']
U = zeros(2,1)
for count = 1:2
U(count) = 0.5* k(count)* d(count)^2
printf("The energy stored in %s is %fJ\n", name(count), U(count))
end
//Sample Problem 10-3b
printf("\n**Sample Problem 10-3b**\n")
for count = 1:2
//Energy conservation
Vf = sqrt(U(count)/(0.5*(m1+m(count))))
//Momentum conservation
Vi = (m1 + m(count))*Vf/m1
printf("The minimum velocity required to break the %s is %fm/s\n", name(count), Vi)
end
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