# 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?

Update:

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?

Relevance

"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!

• 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

• Please provide the rest of the problem.

• Lil1 month agoReport

Never mind I found the rest of the info.