EXTRA CREDIT: A long block of ice is resting on a table. A long trough with para

Question

EXTRA CREDIT: A long block of ice is resting on a table. A long trough with parabolic cross-section is cut out of the ice with length between the focus and the vertex of a 677 A small puck with mass m :0 500 kg is placed a distance h:21 3 cm above the bottom of the trough. What is the horizontal acceleration of the ice block (m#1 7 10 kg) at er the puck is released? Al surfaces are frictionless Express your answer in m s with the-x direction being defned toward the not Note that profs and TAs do not provide help on Extra Credit problems. If you think there is an error in the system, email Enik with your clear reasoning why m2 : CI im

Explanation / Answer

equation of parabola

y = 4ax^2

but a = 6.77 , so,

y = 27.08 x^2

as h = 21.3 cm, using it as y we can get x

21.3 = 27.08 x^2

x = 0.887 cm

Now, slope of the parabola at the point of mass is

dy/dx = 8aX = 54.16 X = 54.16 x 0.887 = 48

but dy/dx = tan@

@ = tan-1 48 = 88.8 deg

Finally, acceleration on a frictionless slope is given by

a = gsin@ = 9.81 sin88.8 = 9.808 m/s^2

Therefore, horizontal acceleration is

ax = acos@ = 9.808cos88.8 = 0.205 m/s^2

Finally,

m1a1 = m2a2 ... By third law

1.71 x a1 = 0.5 x 0.205

a1 = 0.06 m/s^2 towards left

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