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
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?
- NCSLv 710 months 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
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 710 months ago
μ = F/W = 2/9.4 = 0.213
(W+W1)*0.213 = 7.7
W1 = (7.7-9.4*0.213)/0.213 = 26.75 N
- oldschoolLv 710 months ago
Please provide the rest of the problem.