Question
A horizontal force of 200 N is applied to a 55 kg cart across a 10 m level surface. If the cart accelerates at 2 m/s^2 then what is the work done by the force of friction as it acts to retard the motion of the cart? Hint calculate the force of friction using newtons second law.
the answer is -900J...but how do i get that?
the answer is -900J...but how do i get that?
Answers
bobpursley
There are a couple of ways to do this.
Netforce= ma
200N+friction=ma
-200N + ma=friction solve for friction force.
Work done on by friction=forcefricion(10m)
Now notice
Netforce= ma
200N+friction=ma
-200N + ma=friction solve for friction force.
Work done on by friction=forcefricion(10m)
Now notice
thanks
The answer is 900j, since w=fd, we first calculate the net force f=ma=110N then net force - 200N = 90.
Hence: w= (90) (10) = 900j
Hence: w= (90) (10) = 900j
Yes
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