The force applied against a moving object travelling on a linear path is given by #F(x)= sinx + 2 #. How much work would it take to move the object over #x in [ 0, (7 pi) / 8 ] #?

Answer 1

W = int_0^(7/8pi) (sinx+2) dx# # W = ((-cos x)+ 2x)|_0^(7/8pi) # #W = .00115 + 7/4pi #

Work #W = int_a^bF(x) do # #F(x) = sinx + 2; a = 0; b = 7/8pi# # W = int_0^(7/8pi) (sinx+2) dx# # W = ((-cos x)+ 2x)|_0^(7/8pi) # #W = .00115 + 7/4pi #

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Answer 2

Charlotte Aaron

To find the work done, integrate the force function F(x) over the given interval [0, (7π)/8] with respect to x:

∫[0, (7π)/8] (sin(x) + 2) dx

= [-cos(x) + 2x] evaluated from 0 to (7π)/8

= [-cos((7π)/8) + 2((7π)/8)] - [-cos(0) + 2(0)]

= [-cos((7π)/8) + (7π)/4]

≈ 3.595 units of work

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.