Answer to Question #247345 in Mechanics | Relativity for Malvus Nyamadzawo

Question

For a short time the crane in the figure below lifts the 2.50-Mg beam with a force of F = (28+3s2 )kN. Determine the speed of the beam when it has risen s = 3 m. Also, how much time does it take to attain this height starting from rest? [5]

Answer 1

$F=(28+3s^2)[kN]=10^3(28+3s^2)[N] \\ F=ma=m\cfrac{dv}{dt}=m \cfrac{dv}{ds}\cfrac{ds}{dt} \\ \therefore F=mv\cfrac{dv}{ds}=10^3(28+3s^2) \\ \implies \int^{v}_{v=0} vdv=\cfrac{10^3}{m}\int^{s}_{s=0} (28+3s^2)ds \\ \implies \cfrac{1}{2} \bigg[ {v}^2 \bigg]^{v}_{v=0}=\cfrac{10^3\,N}{2500\,kg} \bigg[28s+s^3 \bigg]^{s}_{s=0} \\ \implies \cfrac{v^2}{2} =0.4\cfrac{m}{s^2} \bigg[28s+s^3 \bigg]m \\ \therefore {v} = \sqrt{ 0.8\cfrac{m^2}{s^2} \bigg[28s+s^3 \bigg] }$

Now we substitute to find:

${v} = \sqrt{ 0.8\cfrac{m^2}{s^2} \bigg[28(3)+(3)^3 \bigg] }=9.4234\frac{m}{s}$

Then we find the time with

${v}=\cfrac{ds}{dt} = \sqrt{ 0.8\cfrac{m^2}{s^2} \bigg[28s+s^3 \bigg] } \\ \therefore t=\int^{s}_{s=0} \cfrac{ds}{v} \\ \implies t\approxeq 0.0446429 \log(s) - 0.0223214\log(28 + s^2)$

After substitution (s = 3m), we find:

$t= 0.0446429 \log(3) - 0.0223214\log(28 + 3^2)\approxeq 0.0315 \,s$

Reference:

• Sears, F. W., & Zemansky, M. W. (1973). University physics.