The force acting on a particle varies as in the figure below. (The x-axis is marked in increments of 5.00 m.)A coordinate plane has a horizontal axis labeled x (m) and a vertical axis labeled Fx (N). There are five points, connected by straight lines in order from A to E.
Find the work done by the force as the particle moves across the following distances.
(a) from x = 0 m
to x = 40.0 m
__ J
(b) from x = 40.0 m
to x = 60.0 m
__ J
(c) from x = 0 m
to x = 60.0 m
__J
First, we have to calculate the force along the displacement:
We proceed to find the following equations:
"F_{AB}=\\frac{3x}{10}\n\\\\ F_{BC}=12-\\frac{3x}{10}\n\\\\ F_{CD}=12-\\frac{3x}{10}\n\\\\ F_{DE}=\\frac{3x}{10}-18"
With this information we can find the work done:
"W_{AC}=\\int^C_A F\\,{dx}=\\int^B_A[F_{AB}]\\,{dx}+\\int^C_B[F_{BC}]\\,{dx}\n\\\\ \\text{ }\n\\\\W_{AC} =\\int^{20}_0[3x\/10]\\,{dx}+\\int^{40}_{20}[12-3x\/10]\\,{dx}\n\\\\ \\text{ }\n\\\\W_{AC} =[3x^2\/20]^{20}_0+[12x-3x^2\/20]^{40}_{20}\n\\\\ \\text{ }\n\\\\W_{AC} =[3(20)^2\/20-0]+[12(40-20)-(3\/20)(40^2-20^2)]J\n\\\\ \\text{ }\n\\\\W_{AC} =[3(20)+12(20)-(3)(60)]J=(60+240-180)J=120\\,J"
For the second part:
"W_{CE}=\\int^E_C F\\,{dx}=\\int^D_C[F_{CD}]\\,{dx}+\\int^E_D[F_{DE}]\\,{dx}\n\\\\ \\text{ }\n\\\\W_{CE} =\\int^{50}_{40}[12-3x\/10]\\,{dx}+\\int^{60}_{50}[-18+3x\/10]\\,{dx}\n\\\\ \\text{ }\n\\\\W_{CE} =[12x-3x^2\/20]^{50}_{40}+[-18x+3x^2\/20]^{60}_{50}\n\\\\ \\text{ }\n\\\\W_{CE} =([12(50-40)-(3\/20)(50^2-40^2)]+[-18(60-50)+(3\/20)(60^2-50^2)])J\n\\\\ \\text{ }\n\\\\W_{CE} =(12(10)-(3\/2)(90)-18(10)+(3\/2)(110))J=(120-135-180+165)J\n\\\\ W_{CE}=-30\\,J"
The total work for the whole process will be the sum:
"W_T=W_{AC}+W_{CE}=(120-30)J=90\\,J"
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