Another Question: Find the work done by the kinetic frictional force

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A 300N force is pulling an 90.0 kg refrigerator across a horizontal surface. The force acts at an angle of 22.0° above the surface. The coefficient of kinetic friction is 0.200, and the refrigerator moves a distance of 6.00 m.

i found out the work done by pulling force is 1669 J and its asking for the work done by the kinetic frictional force!
here's what i did but the answer is wrong though...i tried a million times and it's still wrong! i know something's wrong here and i know it's not the acceleration of gravity or m...hmm

i did
mg times mu=9.8*90*.2
then i times it by the distance to get work. 176.4N*6m=1058.4 J
which is WRONG! arghh.......am i missing something here? :rofl:
 

Answers and Replies

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ehild
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zizikaboo said:
A 300N force is pulling an 90.0 kg refrigerator across a horizontal surface. The force acts at an angle of 22.0° above the surface. The coefficient of kinetic friction is 0.200, and the refrigerator moves a distance of 6.00 m.


i did
mg times mu=9.8*90*.2
then i times it by the distance to get work. 176.4N*6m=1058.4 J
which is WRONG!
The friction is equal the normal force multiplied by the coefficient of friction.

[itex] F_{fr} = \mu F_N[/itex]

As the refrigerator slides on a horizontal plane the vertical component of the resultant force is zero. This vertical component is the normal force + vertical component of the pulling force - mg.

[itex] F_N+ F\sin (\alpha) - mg = 0 [/itex].

ehild
 

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