a.A mechanic pushes a 2.8 103 kg car from rest to a speed of v, doing 5500 J of
ID: 1991927 • Letter: A
Question
a.A mechanic pushes a 2.8 103 kg car from rest to a speed of v, doing 5500 J of work in the process. During this time, the car moves 26.0 m. Neglecting friction between car and road, find each of the following.(a) the speed v
b.A 7.0 kg bowling ball moves at 2.00 m/s. How fast must a 2.60 g Ping-Pong ball move so that the two balls have the same kinetic energy?
c.A crate of mass 12.5 kg is pulled up a rough incline with an initial speed of 1.50 m/s. The pulling force is 100 N parallel to the incline, which makes an angle of 16.0° with the horizontal. The coefficient of kinetic friction is 0.400, and the crate is pulled 4.30 m.
How much work is done by gravity?
How much mechanical energy is lost due to friction?
How much work is done by the 100 N force?
What is the change in kinetic energy of the crate?
What is the speed of the crate after being pulled 4.30 m?
d.On a frozen pond, a 7 kg sled is given a kick that imparts to it an initial speed of v0 = 2.4 m/s. The coefficient of kinetic friction between sled and ice is µk = 0.10. Use the work-energy theorem to find the distance the sled moves before coming to rest.
Explanation / Answer
We shall take the value of acceleration due to gravity 'g'=10 m/s^2 Weight of crate = mg= 10x10 = 100 N Component of weight parallel to the inclined plane=mg sin 20=100x0.34 = 34 N Component of weight perpendicular to the inclined plane=mg cos 20=100 x 0.94 = 94 N Normal reaction=Component of weight perpendicular to the inclined plane= 94 N force of friction f = coefficient of kinetic x normal reaction f =0.4 x 94=37.6 N Distance moved 's' = 5 m work done against friction Wf =fs=37.6 x 5= 188 J (a) work done by gravity Wg = -mg sin 20 x s= -34 x5= -170 J ______________________________________… (b)mechanical energy is lost due to friction=Wf=188 J ______________________________________… (c)work done by 100 N force W = 100 x4.30 =430J ______________________________________… (d)The change in kinetic energy= W -Wf -Wg=500-188-170=142 J ______________________________________… (e)initial kinetic energy=(1/2)mu^2=(1/2)10x 2.25=11.25 J final kinetic energy=initialKE+gain in KE=11.25+142=153.25 J final velocity v=sq rt [2KEfinal / mass ]=sq rt [2x153.25 /10] v=sq rt 30.65=5.536 m/s is nearly equal to 5.54 m/s final velocity of crate is 5.54 m/s
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