Question

A force of 40N is re

Answer 1

Sorry the question is: A force of 40N is required to start a 6.0kg box moving across a horizontal concrete dfloor. If 40N force continues the box acceleratesat .90m/s^2. What is the coefficient of kinetic friction?

Answer 2

I already got the coefficient of static friction = .68

Answer 3

^^ Sorry the acceleration should be .80 m/s^2

Answer 4

I thought the answe would be 4.8 since F/m = a and if m is 6.0kg and acceleration is .80m/s^2, plugging it in would give me 4.8N, but solving for a got me .8 instead.

Answer 5

Applied force F = 40N
Uk = coefficientof kinetic friction
Force of Friction Ff = Uk * mg
Fnet = F - Ff = ma
\[40N−Uk (9.8m/s^2 ∗6.0kg)N=6.0kg∗0.80m/s^2 =4.8N\]\[40N−4.8N=U_{k}(9.8m/s^2 ∗6.0kg)\]\[35.2N=U_{k}(9.8m/s^2 ∗6.0kg)\]\[\frac{35.2N}{9.8 m/s^2 * 6.0 kg}=0.5986... \]Correcting for significant figures leaves you with a coefficient of friction of ~0.60

Answer 6

That last part of your answer looked something like this: 35.2N/58.8N=.5986?

Answer 7

Yes :)

Answer 8

Okay thanks.