Question 11 EXTRA CREDIT: A long block of ice is resting on a table. vertex of a

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Question 11 EXTRA CREDIT: A long block of ice is resting on a table. vertex of a=9.27 cm. A small puck with mass m2 = 0.580 kg is placed a distance h-23.9 cm above the bottom of the t ugh. What is the horizontal acceleration of the ice block (m1 2.091 kg) after the Note that profs and TAs do not provide help on Extra Credit problems. If you think there is an error in the system, email Erik with your clear reasoning why. long trough with parabolic cross-section is cut out of the ice with length between the focus and the Not complete Marked out of 10.0 P Flag questlon is released? All surfaces are frictionless. Ex answer n m2 : a m1 Answer:

Explanation / Answer

equation of parabola

y = 4ax^2

but a = 9.27 , so,

y = 37.08 x^2

as h = 23.9 cm, using it as y we can get x

23.9 = 37.08 x^2

x = 0.8cm

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

dy/dx = 8aX = 74.16 X = 74.16 x 0.8 = 59.328

but dy/dx = tan@

@ = tan-1 59.328 = 89 deg

Finally, acceleration on a frictionless slope is given by

a = gsin@ = 9.81 sin89 = 9.808 m/s^2

Therefore, horizontal acceleration is

ax = acos@ = 9.808cos89 = 0.1712 m/s^2

Finally,

m1a1 = m2a2 ... By third law

2.901 x a1 = 0.58 x 0.1712

a1 = 0.03423 m/s^2

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