# What is the weight of this block?

A weight W is now placed on the block and

7.7 N is needed to push them both at constant

velocity.

What is the weight W of the block?

Answer in units of N.

So there was a part one to this problem. Here it is. Sorry about the confusion.

A block weighing 9.4 N requires a force of 2 Nto push it along at constant velocity.What is the coefficient of friction for thesurface?

### 3 Answers

- NCSLv 71 month agoFavorite Answer
"constant velocity" means "equilibrium," and so the applied horizontal force must equal the friction force.

So from the initial problem, we found from

friction force = µ * normal force

that

µ = 2N / 9.4N = 0.21

and so for the current problem

friction force = µ * new normal force

7.7 N = 0.21 * (W + 9.4N)

which solves to

W = 27 N

as the answers should have no more than 2 significant digits (one might argue for 1 significant digit owing to the data "2 N.")

Hope this helps!

- oubaasLv 71 month ago
a)

μ = F/W = 2/9.4 = 0.213

b)

(W+W1)*0.213 = 7.7

W1 = (7.7-9.4*0.213)/0.213 = 26.75 N

Never mind I found the rest of the info.