As shown in the figure below, a light string that does not stretch changes from

Question

As shown in the figure below, a light string that does not stretch changes from horizontal to vertical as it passes over the edge of a table. The string connects m1, a 3.80 kg block, originally at rest on the horizontal table at a height 1.25 m above the floor, to m2, a hanging 3 kg block originally a distance d = 0.970 m above the floor. Neither the surface of the table nor its edge exerts a force of kinetic friction. The blocks start to move from rest. The sliding block m1 is projected horizontally after reaching the edge of the table. The hanging block m2stops without bouncing when it strikes the floor. Consider the two blocks plus the Earth as the system.

Explanation / Answer

Part (a)
Tension T pulling m1 to the right gives
(1) ..... T = m1 * a
Forces acting on m2 moving downward give
.......... T + ( - m2 * g ) = m2 * ( - a )
(2) ..... T = m2 * g - m2 * a
Substituting (1) in (2) gives
.......... m1 * a = m2 * g - m2 * a
.......... ( m1 + m2 ) * a = m2 * g
(3) ..... a = ( m2 * g ) / ( m1 + m2 )
m2 moving downward from rest with acceleration given in (2) yields
.......... 0.960 m = d = ( 0.5 ) * a * ( t^2 )
.......... ( t^2 ) = 2 * d * ( m1 + m2 ) / ( m2 * g )
(4) .. ... t = sqrt [ 2 * d * ( m1 + m2 ) / ( m2 * g ) ]
Substituting values in (4) gives t = 0.66 s
m1 moving to the right from rest with acceleration in (3) and time given
given in (4) attains final speed of
.......... vx = a * t = [ ( m2 * g ) / ( m1 + m2 ) ] * sqrt [ 2 * d * ( m1 + m2 ) / ( m2 * g ) ]
(5) ..... vx = sqrt [ ( 2 * d * m2 * g ) / ( m1 + m2 ) ]
after which m1 moves with uniform velocity on frictionless table top since
m2 has landed on the ground.
The speed at which m1 leaves the edge of the table is that given in (5).
Substituting the values, we get vx = 2.92 m/s
ANSWER IN (a) IS 2.92 m/s

Part (b)
Without any acceleration along the horizontal, the x-component of velocity
of m1 during its fall remains constant at 2.92 m/s.
The y-component of velocity increases from zero to some value vy given by
(6) ..... vy^2 = 2 * g * h
where h = 1.27 m is the height of the table top.
The magnitude of the velocity with which m1 hits the ground is therefore
(7) ..... v = sqrt [ vx^2 + 2 * g * h ]
Substituting the values in (7) gives v = 5.78 m/s
ANSWER IN (b) IS 5.78 m/s

Part (c)
Since the y-component of velocity of m1 is zero when it leaves the table top, the
time of fall of m1 can be obtained from 1.27 m = h = ( 0.5 ) * g * ( t^2 ) which gives
......... t = sqrt [ 2 * h / g ] = 0.51 s
The horizontal distance traveled by m1 in that time is
......... S = vx * t = ( 2.92 ) * ( 0.51 ) = 1.49 m
ANSWER IN (c) IS 1.49 m.

sorry i have just done the similar type of problem please plugin the numbers if any problem just comment on this

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