The force applied against a moving object travelling on a linear path is given by #F(x)= x^2+e^x #. How much work would it take to move the object over #x in [0, 5] #?

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Answer 1 · 2017-10-26T07:52:14

see below

for a variable force

Work done#=int_a^b(F(x)dx#

in this case

Work done#=int_0^5(x^2+e^x)dx#

#=[1/3x^3+e^x]_0^5#

#=[1/3x^3+e^x]^5-[1/3x^3+e^x]_0#

#=1/3 5^3-e^5-0-e^0#

#=125/3-1-e^5#

#=(122/3-e^5)J#