A box of mass m=l.Okg is at rest at point A. AB is a friction less plane incline

Question

A box of mass m=l.Okg is at rest at point A. AB is a friction less plane inclined at 30degree with respect to the horizontal. BCD is a horizontal plane. At point B there is a smooth transition between the motion on the two planes (i.e. ms speed just before arriving at B is equal to the horizontal speed just after leaving B). At point C the mass makes contact with an uncompressed spring of spring constant k 480N/m. C is 4m from B. The coefficient of kinetic friction on the segment from B to C is 0.15. What is the acceleration (in m/s^2) of the box at point A? If the box takes 3s to travel from A to B. what is its speed (in m/s) at point B? What is the work (in Joules) that gravity docs on the box during the motion from A to B? What is the kinetic energy (in Joule) of the box at B? What is the work (in Joule) that friction does on the box during the motion from B to C? What is the kinetic energy (in Joule) of the box at C? In order to bring the box to a full rest, by how much does the length of the spring get compressed

Explanation / Answer

part a

accelartion of block at A = gsin30 = g/2= 9.81/2= 4.925 m/s2

part b

t= 3 sec

vi= 0 m/s

so by kinematics first equation

vb = vi+ at = 0+(gsin30)*3 = 14.715m/s

pART C

total distance coverd from A-B = dab

by kinematics second equation

S= vi*t + 0.5at2= 0 +0.5*4.905*32 = 22.0725 m

so work done = force in direction of displacement * displacement= mgsin30*22.0725 = 1*9.81*sin30*22.0725 = 108.2656 J

part d

by law of energy coservation

work done by garvity = kinetic energy

so

K.E.at B = 108.2656J

part e

work done by friction = friction * dbc

friction = 0.15*mg= 0.15*1*9.81=1.4715 N

work done by frictioin = 1.4715*4=5.886 J

part f

so K.E at c = K.E at B - workdone by friction on block m=108.2656J -5.886 J = 102.3796 J

part g

K.Eat C = 0.5kx2= 102.3796

put valuve of K and solve

0.5*480*x2= 102.3796

x= 0.6531 m

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