How is the Work Done by Friction Calculated on a Box?

[snoopyrawr]

Homework Statement

i've been stuck on this problem for about 45mins.

A constant external force P = 160 N is applied to a 20 kg box,which is on a rough horizontal surface. The force pushes the box a distance of 8.0 m, in a time interval of 4.0 s, and the speed changes from v1 = 0.5 m/s to v2 = 2.6 m/s. Find the work done by friction. The work done by friction is closed to:
There is a force applied at 30degree to the horizontal of the box.

the answer is -1043 J but idk how to get that answer.

Homework Equations

Ff = µFn
W=Fd
W=mgd

The Attempt at a Solution

W=mgd
W=20kg*9.8*cos(30)*8=1357.9J

 

[tiny-tim]

welcome to pf!

hi snoopyrawr! welcome to pf!

A constant external force P = 160 N is applied to a 20 kg box,which is on a rough horizontal surface. The force pushes the box a distance of 8.0 m, in a time interval of 4.0 s, and the speed changes from v1 = 0.5 m/s to v2 = 2.6 m/s. Find the work done by friction.

the answer is -1043 J but idk how to get that answer.
…
W=mgd
W=20kg*9.8*cos(30)*8=1357.9J

(so the force is applied at 30° to the horizontal?)

W = mgd is wrong

W = Pd is right

(and anyway, that's the work done by the force P … the question asks for the work done by the friction )

try again ​

 

[snoopyrawr]

yes there is a horizontal force of 30degrees applied on the box and how would you set up this problem if the friction coefficient is not given?

 

[tiny-tim]

use the work energy theorem

 

[snoopyrawr]

so it would be like this:
W=KEf-KEi (KEf= kinetic final, KEi= kinetic initial)
W=(1/2(20kg)(2.6m/s)^2)-(1/2(20kg)(0.5m/s)^2)
W=65.1 J

is there another step i have to do?

 

[tiny-tim]

(just got up :zzz:)

so it would be like this:
W=KEf-KEi (KEf= kinetic final, KEi= kinetic initial)
W=(1/2(20kg)(2.6m/s)^2)-(1/2(20kg)(0.5m/s)^2)
W=65.1 J

yup!

is there another step i have to do?

that's the total work done …

so what's the work done by friction? ​