ecoli Posted December 21, 2004 Posted December 21, 2004 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.
5614 Posted December 21, 2004 Posted December 21, 2004 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.
bloodhound Posted December 21, 2004 Posted December 21, 2004 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)
bloodhound Posted December 21, 2004 Posted December 21, 2004 Friction = 10N C of F = 0.203 (to 3 decimal places)
[Tycho?] Posted December 22, 2004 Posted December 22, 2004 Oh I didn't get the mu, didn't know the question wanted it, oops.
fuhrerkeebs Posted December 22, 2004 Posted December 22, 2004 40+Frictional force=6*5, Frictional Force=10. N=g*5~49. 49*coef. of friction=10, coef. of friction~.204.
ecoli Posted December 22, 2004 Author Posted December 22, 2004 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.
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