Another physics question

[pizza1994]

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How much work is done by the normal force as a 13 kg penguin slides down an iceberg 40 m long at an incline of 45° to the horizontal?

1) O J

b) 1850 J

c) 3700 J

d) 5200 J

any help would be appreciated guys!!!! 🙂

 

[el_duderino]

W = fd
Distance is 40/sqrt(2) = 28.28 m
Force = 13 kg * 9.8 m/s^2 = 127.4 N

28.28 m * 127.4 N = 3602 J

Pretty sure that's right....

 

[Rhino1000]

W = fd
Distance is zero
Force = 13kg*9.8m/s^2 = 127.4N
0 * 127.4N = 0 J.

Zero work is done because the normal force is perpendicular to the slope and therefore the direction of the objects movement.

Work is the amount of force multiplied by the distance travelled along the force's vector. Or the displacement multiplied by the force component parallel to the displacement.

 

[el_duderino]

Oh, wait. Work done by the NORMAL force. Derp. Hm. I think that would have to be zero, as there is no work done in the direction of the normal force. The direction of translation is exactly perpendicular to the normal force.

 

[pizza1994]

ahh so everytime they ask for the work done by the normal force then its always zero??? cause the normal force is always perpendicular?

 

[el_duderino]

Yeah, I think so. I'm trying to imagine a situation where it wouldn't, but can't.

 

[DrknoSDN]

ahh so everytime they ask for the work done by the normal force then its always zero??? cause the normal force is always perpendicular?

Yeah, I think so. I'm trying to imagine a situation where it wouldn't, but can't.

A ball that bounces off the floor would have Work equal to mgh that is produced by the normal force.
The force is applied in the same direction as the velocity vector.

 

[NextStepTutor_1]

Similarly, the normal force in an elevator acting on you as the elevator moves upwards does work since the force and displacement are in the same direction.