Question

Work done on this block.

Answer 1

How much work is need to move a block that is 4.00Kg over a 0.025m distance?

If the force required to over come static friction is 7.84N
coefficient of static friction is: 0.2
coefficient of kinetic friction = 0.1
4.00Kg x 9.8 = 39.2 x 0.2 =7.84N
39.2 x 0.1 = 3.92N

So the amount of force needed to move a block in that direction is 8N would be enough?
Work needed: 0.2J?

Is my calculation right here?
+ Is it even possible to have static friction = kinetic friction?

Answer 2

@Jemurray3 @Vincent-Lyon.Fr @TuringTest @ghazi @Shane_B

Answer 3

@experimentX @.Sam. @amistre64 @ash2326

Answer 4

Hi @experimentX
So is my calculation and my solution correct?

Answer 5

don't know how to work with two friction coefficients. from where to where (up to what velocity) do you consider static friction and from where to where (velocity) do you consider kinetic friction?

Answer 6

Hmm, Im not sure myself.

Answer 7

I mean, static friction is that friction at the beginning...

Answer 8

K.friction is the one that starts as soon as the object starts to move.

Answer 9

I assume, I should be only work on K.friction

Answer 10

But generally, how would you solve this problem?
How much work do you need(minimum) to start moving a block of 4kg 0.025m?

Answer 11

i would use Kinetic friction. No work is done with static friction since the body is not moving.

Answer 12

So the force needed to move this object must be greater than 3.92
So, work = 4N x 0.025m = 0.1 J?

Answer 13

Hi @theEric
What do you think?

Answer 14

I just got to the bottom. I'm going to look at the question again and reply! :)

Answer 15

Assuming the body possess velocity from beginning and constant force is being applied to keep it in uniform motion.
\[ 4*9.8*0.1*0.025 \approx 0.1 J \]

Answer 16

So, I would agree that you only need kinetic friction. Work would be \(W=F\ d\ \cos(\theta)\) since the force used to move the box is directed in the same direction as the distance. The force to overcome static friction is applied for no distance, really...

So, the force you use to push is opposite to friction. So \(W=-F_{f_\text k}\ d\), and \(F_{f_\text k}=m\ g\ \mu_\text k\).

Answer 17

I'm assuming it has constant velocity too :)

Answer 18

That is the minimum work that can be done.

Answer 19

Yes it would have a constant force.
So I would require at least 0.1J to move that block correct :D ?

Answer 20

Thanks!

Answer 21

yes

Answer 22

You're welcome! :)

Answer 23

I agree with the 0.1 J derived by experimentX