(A)A 66.0 kg base runner begins his slideinto second base when he is moving at a
ID: 1764674 • Letter: #
Question
(A)A 66.0 kg base runner begins his slideinto second base when he is moving at a speed of 3.6 m/s. The coefficient of friction between hisclothes and Earth is 0.70. He slides so that his speed is zero justas he reaches the base. (1) How much mechanical energy is lost due tofriction acting on the runner?1 J
(2) How far does he slide?
2 m
(B)A 7.80-g bullet moving at 460 m/spenetrates a tree trunk to a depth of 4.4cm. (1) Use work and energy considerations to findthe average frictional force that stops the bullet.
1 N
(2) Assuming the friction force is constant, determine how muchtime elapses between the moment the bullet enters the tree and themoment it stops moving.
2 (1) How much mechanical energy is lost due tofriction acting on the runner?
1 J
(2) How far does he slide?
2 m
(B)A 7.80-g bullet moving at 460 m/spenetrates a tree trunk to a depth of 4.4cm. (1) Use work and energy considerations to findthe average frictional force that stops the bullet.
1 N
(2) Assuming the friction force is constant, determine how muchtime elapses between the moment the bullet enters the tree and themoment it stops moving.
2 (1) Use work and energy considerations to findthe average frictional force that stops the bullet.
1 N
(2) Assuming the friction force is constant, determine how muchtime elapses between the moment the bullet enters the tree and themoment it stops moving.
2
Explanation / Answer
(a) Given that the mass of runner is m = 66kg Initial velocityis U = 3.6 m/s he coefficient offriction between his clothes and Earth is = 0.70 ------------------------------------------------------------------------------------- The mechanical energy lost duringmotion is K = (1/2)mU2 = (1/2)(66kg)(3.6m/s)2 = 427.68 J From the work energy theoremthis kinetic energy = work done due to friction 42.7.68J = f *d 42.7.68J = mg*d d = 42.7.68J / mg =----------- m (b) Given that the initial velocity is U = 460 m/s mass of body is m =7.80*10-3kg Distance traveled is d = 0.044 m From the work energy theorem this kineticenergy = work done due to friction (1/2)mU2 = f *d Then the frictional force is f =(1/2)mU2 / d = -------- N But The deceleration due to thefrictional force is a = - f /m From the equation of motion timetaken is t = (V - U) / a = ( 0- U) /a = - U /(-f / m) = mU /f =--------- sec
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