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Workenergy theorem problem  I have it close to correct 
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#1
Oct2309, 06:32 PM

P: 190

1. You are a member of an alpine rescue team and must project a box of supplies, with mass m, up an incline of constant slope angle alpha so that it reaches a stranded skier who is a vertical distance h above the bottom of the incline. The incline is slippery, but there is some friction present, with kinetic friction coefficient mu_k.
Use the workenergy theorem to calculate the minimum speed v that you must give the box at the bottom of the incline so that it will reach the skier. Express your answer in terms of some or all of the variables m, g, h, mu_k, and alpha. 2. K_final = 0 K_initial = 0.5mv^2 net work = F(delta r)*cos(theta) 3. Normal force = mg Force in x direction = mg[mu_k  sin(alpha)] net work = K_final  K_initial = 0.5mv^2 = F(delta r)*cos(alpha) = mg[mu_k + sin(alpha)]*h*cos(alpha) finally: v = sqrt[2gh(mu_k + sin(alpha))*cos(alpha)] Apparently my trigonometry is wrong (i.e. I have a cos and/or sin mixed up), but I am not sure where my error is. 


#2
Oct2309, 08:43 PM

HW Helper
P: 3,394




#3
Oct2309, 09:00 PM

P: 190

Oh, duh. But this is still incorrect.
With force in the x direction now = mg[mu_k*cos(alpha)  sin(alpha)], net force = 0.5mv^2 = mg[mu_k*cos(alpha)  sin(alpha)]*h*cos(alpha) v = sqrt(2gh[mu_k*cos(alpha) + sin(alpha)]*cos(alpha)) What am I doing wrong. Alternatively, I can get this using kinematics: v_final^2 = v_initial^2 2a(delta s) v_i = sqrt(2a*delta s) v_i = sqrt[2*(g(mu_kcos(alpha)  sin(alpha))) * h] is this correct? If so where did I go awry in the first method? Thank you. 


#4
Oct2309, 09:11 PM

HW Helper
P: 3,394

Workenergy theorem problem  I have it close to correct
Okay on F = mg[mu_k*cos(alpha)  sin(alpha)] along the hill.
Then you want Ek = F*d, where the distance along the hill is d = h/sin(A). I don't follow your other way. What is "s"? My other way is Ek = work lost to friction + mgh. It gives the same answer (with a cos(A)/sin(A) in it from the friction term). 


#5
Oct2309, 09:26 PM

P: 190

In the other method s is just displacement.
Ok, so Ek = mg(mu_kcosA  sinA) = Fd Why is it Fd? and is d equivalent to delta r? So you get: [mgh(mu_kcosA  sinA)*cosA]/sinA = mgh(mu_kcosA  sinA)*cotA Is that really my answer? The cot just seems intuitively funny to me. 


#6
Oct2309, 09:49 PM

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P: 3,394

Maybe we should throw away all those minus signs!
F = mg(mu_kcosA + sinA) Both forces are downward along the ramp but we have to push upward to overcome them. I wrote W = Fd because you had the force negative. ½mv² = Fs = mg(mu_kcosA + sinA)*h/sin(A) You seem to have an extra cosA in yours. 


#7
Oct2309, 09:53 PM

P: 190

I had that extra cos because of net work = F(delta r) cosA
thank you. 


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