An object with a mass of 5 kg5kg is pushed along a linear path with a kinetic friction coefficient of u_k(x)= x+xsinx uk(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

7 "kJ"7kJ

Explanation:

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

The friction is given by F = u_"k" * NF=ukN.

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

The total work done to oppose the friction is given by

int_0^{5pi} F dx = N int_0^{5pi} u_k(x) "d"x5π0Fdx=N5π0uk(x)dx

= 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"