Could anyone help me with this problem?

 

 

a force of 40.0N accelerates a 5.0kg block at 6.0m/s^2 along a horizontal surface. How large is the frictional force? What is the coefficient of friction?

 

I can't figure out how to get the force of friction without given the proportionality constant.

surely the friction would count as the resistive force in this case and you could effectively ignore air resistance?

 

what formula were you given to solve the problem or werent you?

 

if not, i would go with friction = all resistive force

 

which should fit will into an equation.

first ... net force = mass times acceleration. applied force - net force = resistive force.

 

assuming resistive force = frictional thingy

 

frictional = coef of friction times the Normal force (reaction force)

Its 10 I think.

Friction = 10N

 

C of F = 0.203 (to 3 decimal places)

Oh I didn't get the mu, didn't know the question wanted it, oops.

40+Frictional force=6*5, Frictional Force=10. N=g*5~49.

49*coef. of friction=10, coef. of friction~.204.

applied force - net force = resistive force.

 

Thanks, that's the equation I was missing. I couldn't get from applied force to Force of Friction (the only resistive force being considered).

Thanks for your help guys.