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Work done by friction on a box

  • Thread starter snoopyrawr
  • Start date
1. 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.

2. Homework Equations
Ff = µFn
W=Fd
W=mgd

3. The Attempt at a Solution
W=mgd
W=20kg*9.8*cos(30)*8=1357.9J
 
Last edited:

tiny-tim

Science Advisor
Homework Helper
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246
welcome to pf!

hi snoopyrawr! welcome to pf! :smile:
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 :wink:)

try again :smile:
 
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

Science Advisor
Homework Helper
25,790
246
use the work energy theorem
 
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

Science Advisor
Homework Helper
25,790
246
(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! :smile:
is there another step i have to do?
that's the total work done …

so what's the work done by friction? :wink:
 

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