An object with a mass of #6 kg# is pushed along a linear path with a kinetic friction coefficient of #u_k(x)= 1+5cotx #. How much work would it take to move the object over #x in [(7pi)/12, (5pi)/8], where x is in meters?

1 Answer
Nov 27, 2017

Answer:

#W = \int_{(7\pi)/12}^{(5\pi)/8} F_{f}(x).dx = \pi/24 + 5\ln{\sin((5\pi)/8)/\sin((7\pi)/12)}#

Explanation:

#\mu_k(x) = 1 + 5\cotx; \qquad F_{f} (x) = \mu_k(x)mg; #

#W = \int_{(7\pi)/12}^{(5\pi)/8} F_{f}(x).dx#
#= \int_{(7\pi)/12}^{(5\pi)/8}(1+5\cotx)dx = [x + 5( \ln(|\sinx|) ) ]_{(7\pi)/12}^{(5\pi)/8} #
#=((5\pi)/8 - (7\pi)/12) + 5\times{\ln[\sin((5\pi)/8)] - \ln[\sin((7\pi)/12)]}#
#= \pi/24 + 5\ln{\sin((5\pi)/8)/\sin((7\pi)/12)}#