PH111 - MECHANICS
Solution to Quiz 9

Thursday 10/29/15	Name
O101: periods 1,2
	CM#

A 5.00 kg block moves along a smooth, horizontal floor. The applied force is

F=(20.0 N)Ð38.0o

F=ma P=F·v vf=vi+at Pav=F·vav

(a) Calculate the acceleration of the block.



ax = Fcosq m = (20.0 N)cos38.0o 5.00 kg = 3.15 m/s2


(b) If the block starts from rest, calculate the speed of the block at t=4.00 s.



vf = vi+axt = 0 + (3.15 m/s2)(4.00 s) = 12.6 m/s


(c) Calculate the instantaneous power delivered to the block at t=4.00 s



P = F·v = Fvfcosq = (20.0 N)(12.6 m/s)cos38.0o = 199 W


(d) Calculate the average power delivered to the block during the time interval 0 £ t £ 4.00 s.



Pav = F·vav = Fvavcosq = (20.0 N)(6.30 m/s)cos38.0o = 99.3 W

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On 05 Nov 2015, 11:46.