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

• Amber430
In summary, a 46 kg box is being pushed a distance of 7.0 m across the floor by a force vector P with a magnitude of 148 N, parallel to the displacement of the box. The work done on the box by the applied force is 1036 J. For the frictional force, Fk= 0.25 x Fn, and the work done is -789 J. The normal force does no work as it is completely countered by the force of gravity, which also does no work. The work done by gravity is 0 J.
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.

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.

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.

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

## 1. How do you calculate the work done on a 46 kg box by applied forces?

The work done on a 46 kg box by applied forces can be calculated using the formula W = Fd, where W is the work done, F is the applied force, and d is the distance the box is moved in the direction of the force.

## 2. What is the role of frictional force in the work done on a 46 kg box?

Frictional force plays a role in the work done on a 46 kg box by slowing down the motion of the box. This means that the work done by the applied force has to overcome the force of friction, resulting in a decrease in the overall work done on the box.

## 3. How does the normal force affect the work done on a 46 kg box?

The normal force, which is the force exerted by a surface on an object in contact with it, does not contribute to the work done on a 46 kg box. This is because the normal force acts perpendicular to the direction of motion, so it does not do any work.

## 4. Does the weight of the 46 kg box affect the work done on it?

Yes, the weight of the box, which is the force of gravity acting on it, does contribute to the work done on the box. This is because the weight is an applied force that is acting in the same direction as the motion of the box, so it adds to the overall work done.

## 5. How does the direction of the applied force affect the work done on a 46 kg box?

The direction of the applied force affects the work done on a 46 kg box by determining whether the force is doing positive or negative work. If the force is acting in the same direction as the motion of the box, it does positive work. If it is acting in the opposite direction, it does negative work.

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