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jhae2.718
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Homework Statement
An object of mass m=200g slides down a frictionless ramp from a height H= 60 cm. Near the bottom of the ramp, the path changes into an arc of a circle with radius r= 20 cm. The arc has coefficient of kinetic friction μk=0.4. The angle θ=120° and points P1 and P2 which are the ends of the arc, are at the same height.
Questions:
*What is the speed of the object at point P1?
*What is the kinetic energy of the object at point P2?
Homework Equations
[itex]W_{ext}={\Delta}U+{\Delta}K+{\Delta}E_{therm}
[/itex]
[itex]
{\Delta}E_{therm}=F_ks_{rel}
[/itex]
[itex]
K=\frac{mv^2}{2}
[/itex]
[itex]
U=mgh
[/itex]
[itex]
F=ma
[/itex]
The Attempt at a Solution
Finding the speed at point P1 is done through a simple application of the Work-Energy Theorem (letting U0 be at the bottom of the arc):
[itex]
{\Delta}H=H-r+r\sin{60^{\circ}}
[/itex]
[itex]
mg{\Delta}H=\frac{mv^2}{2}
[/itex]
[itex]
v=\sqrt{2g{\Delta}H}
[/itex]
[itex]
v=\sqrt{2\left(9.80665 \,\mathrm{m/s^2} \right)\left(0.4+0.2\sin{60^{\circ}} \,\mathrm{m} \right)}
[/itex]
[itex]
v \approx 3.35298\,\mathrm{m/s}
[/itex]
Finding the kinetic energy at P2 is much more difficult. Here, the surface transforms from a frictionless one to one with coefficient of kinetic friction 0.4. Since the height of P1 is the same as at P2, ΔU=0. Furthermore,
[itex]
{\Delta}K=\frac{mv_f^2}{2}-\frac{mv_0^2}{2}
[/itex]
and
[itex]
{\Delta}E_{therm}=F_ks_{rel}=\mu_kF_ns_{rel}
[/itex]
The problem is that Fn is dependent on the angle at which the force of gravity acts relative to the surface of the curve. Additionally, the net force on the object is equal to the centripetal force on the object, [itex]\frac{mv^2}{r}[/itex]. Thus, to calculate the force of friction it becomes necessary to account for a variable force which is dependent on position and velocity. Velocity is also dependent on the force of kinetic friction.
The normal force is
[itex]
F_n=mg\cos{\phi}+\frac{mv^2}{r}
[/itex], where phi is the angle between the force of gravity and its component parallel to the normal force. Thus, the thermal energy is equal to:
[itex]
{\Delta}E_{therm}=F_ks_{rel}=\mu_k \left( mg\cos{\phi}+\frac{mv^2}{r} \right) s_{rel}
[/itex]
The total distance srel is the arc length, [itex]\frac{2\pi r}{3}[/itex].
Thus, the problem becomes deriving an expression for Fk in terms of position or angle which can be integrated to find the work done by the force.
I've tried writing a calculator program in TI-Basic to find the force at different positions, but it hasn't yielded the answer.
U is μk, M is the mass of the block, G is acceleration due to gravity, X is the angle, and R is the radius of the curve. Z represents Fn at the point, and T is just used to start the list.
Code:
For (X,-2pi/3,2pi/3,.01)
U(MGcos(X)+MV^2/R)(XR)->{Z}
sqrt(V^2-Z)->V
If T=0
Then
{Z}->L1
1->T
Else
augment({Z},L1)->L1
End
End
The book says the answer is 0.71 J.
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