A toboggan approaches a snowy hill moving at 13.3 m/s . The coefficients of static and kinetic friction between the snow and the toboggan are 0.430 and 0.310, respectively, and the hill slopes upward at 45.0 ∘ above the horizontal. Part A Find the acceleration of the toboggan as it is going up the hill. Assume +x axis directed up the hill. ax = m/s2 SubmitMy AnswersGive Up Part B Find the acceleration of the toboggan after it has reached its highest point and is sliding down the hill. Assume +x axis directed up the hill.
force acting on toboggan while going upwards (along - x )= mu*mgcos(theta) + mgsin(theta)
acceleration = mu*gcos(45) + gsin(45) = -9.08709995571 m/s2
negative as acceleration is downhill
as it reaches highest point
force downhill = mgsin(45)
force uphill = mu*mgcos(45)
acceleration = gsin(45)- mu*gcos(45) = -3.9539 m/s2 ax
since it is coming positive means
A toboggan approaches a snowy hill moving at 13.3 m/s . The coefficients of static and...
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Potential Energy and Force A mass m=3.00kg moving along the x-axis is acted on only by a single conservative force. The force has a potential energy function given by U(x)=(1.00J/m3)x3−(9.00J/m2)x2+(15.0J/m)x. It will be useful to graph this function on your calculator or computer. Part A Find the force on the mass as a function of x. (Leave the units out of the coefficients in your expression, and make sure all coefficients have three significant figures). F(x) = N SubmitMy AnswersGive...
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