Work Done on 46 kg Box by Forces: Applied, Frictional, Normal, Gravity

[Amber430]

A 46 kg box is being pushed a distance of 7.0 m across the floor by a force vector P whose magnitude is 148 N. The force vector P is parallel to the displacement of the box. The coefficient of kinetic friction is 0.25. Determine the work done on the box by each of the four forces that act on the box. Be sure to include the proper plus or minus sign for the work done by each force.

applied force - 1036 J (got this one right)

frictional force (J)-

normal force (J)-

gravity (J) -

1) I got the applied force right. For frictional force, Fk= 0.25 x Fn. I though the normal force equaled W, and W= 9.8 x 46 kg= 451 N. So for frictional force I got 113 N. Since it needs to be in J, I thought it had to be multiplied by the distance, 7 m, so I multiplied 113 by 7, but the answer is wrong.

2) For normal force, it also needs to be in J so I multiplied 451 by 7. The answer is wrong.

3) For gravity, not sure what to do.

 

[Amber430]

Okay I figured out frictional force, it is -789 J. I just forgot the negative sign. But for normal force, wouldn't it be the same as W (46 kg x 9.8= 451 N)? This is the value I plugged into the frictional force equation...but when I multiply 451 by 7 and plug in the answer for normal force it's wrong.

 

[Delphi51]

The normal force does no work. The object does not move in the direction of the normal force (down), so W = F*0. In fact, this force of gravity is completely countered by the force of the table pushing up on the object. That's why it doesn't accelerate up or down.

 

[Amber430]

Ohhh okay! I get it now! I feel dumb now, Lol. Thank you so much