An object with a mass of #5 kg# is pushed along a linear path with a kinetic friction coefficient of #u_k(x)= e^x+x #. How much work would it take to move the object over #x in [1, 2], where x is in meters?

2 Answers
May 7, 2017

Answer:

283.53J

Explanation:

First we will write the information provided
#mu_k=e^x+x#
M=#5 Kg#

we know that the work done by us to the 5 kg mass is
#W=F_(applied)*ds#

but in this case there is friction is involved which can be included in the equation by ,

#W=(F_(applied)-f_(k))*dS#

here we need to move the block without acceleration so the applied force must be equal to the kinetic friction .
#F_(applied)=f_k#

similarly the work done will also be the same in both the cases so the work done by the frictional force is given by .

#W=f_k*dx=mu_k*N*dx=int_1^2(e^x+x)*m*g*dx=283.53J#

May 7, 2017

Answer:

The work is #=302.3J#

Explanation:

The work done is

#W=F*d#

The frictional force is

#F_r=mu_k*N#

#N=mg#

#F_r=mu_k*mg#

#=5(e^x+x)g#

The work done is

#W=5gint_(1)^(2)(e^x+x)dx#

#=5g*[e^x+x^2/2]_(1)^(2)#

#=5g((e^2+2)-(e+1/2)))#

#=5g(9.39-3.22)#

#=5g*6.17#

#=302.3J#