An object with a mass of #5 kg# is pushed along a linear path with a kinetic friction coefficient of #u_k(x)= x+xsinx #. How much work would it take to move the object over #x in [0, 5pi], where x is in meters?

1 Answer
Mar 13, 2016

Answer:

#7 "kJ"#

Explanation:

The weight of the object is given by #W = m*g#. Since the object is not accelerating in the vertical direction, the normal force, #N#, is exactly equal to the weight.

The friction is given by #F = u_"k" * N#.

Consider the object being dragged/pushed along a small distance #"d"x#. The amount of work needed is #F * "d"x = N u_k(x) "d"x#.

The total work done to oppose the friction is given by

#int_0^{5pi} F dx = N int_0^{5pi} u_k(x) "d"x#

#= mg int_0^{5pi} (x+xsin(x)) "d"x qquad# (integrate by parts)

#= mg [x^2/2 - xcos(x) + sin(x)]_0^{5pi}#

#= mg (frac{25pi^2}{2} + 5pi)#

#= (5 "kg") (9.81 "m/s"^2) (frac{25pi^2}{2} + 5pi)#

#= 6821 "J"#