Comments on: Can someone helpme out with Potential Energy and Conservation Energy question? http://www.energy--talk.com/alternative-energy/can-someone-helpme-out-with-potential-energy-and-conservation-energy-question/97/ How Can We Be More Energy Efficient? Thu, 09 Feb 2012 00:39:04 +0000 http://wordpress.org/?v=2.6.5 By: gp4rts http://www.energy--talk.com/alternative-energy/can-someone-helpme-out-with-potential-energy-and-conservation-energy-question/97/#comment-191 gp4rts Sat, 10 Nov 2007 06:35:49 +0000 http://www.energy--talk.com/alternative-energy/can-someone-helpme-out-with-potential-energy-and-conservation-energy-question/97/#comment-191 The initial energy is 0.5*m*v². The energy used in climbing the ramp is potential energy of the elevation increase = m*g*h plus the energy of friction = Fr*L. L is the distance it travels up the ramp. Also, h/L = sinθ, so the potential energy is m*g*L*sinθ. Fr is the frictional force, which is µ*Fn, where Fn is the normal component of the weight on the ramp. Fn = m*g*cosθ, so Fr = µ*m*g*cosθ and frictional energy is µ*m*g*L*cosθ These energies added together must equal the initial kinetic energy: 0.5*m*v² = m*g*L*sinθ + µ*m*g*L*cosθ solve for L: 0.5*v² = g*L*(sinθ + µ*cosθ) L = 0.5*v²/[ g*(sinθ + µ*cosθ)] The initial energy is 0.5*m*v². The energy used in climbing the ramp is potential energy of the elevation increase = m*g*h plus the energy of friction = Fr*L. L is the distance it travels up the ramp. Also, h/L = sinθ, so the potential energy is m*g*L*sinθ. Fr is the frictional force, which is µ*Fn, where Fn is the normal component of the weight on the ramp. Fn = m*g*cosθ, so Fr = µ*m*g*cosθ and frictional energy is µ*m*g*L*cosθ These energies added together must equal the initial kinetic energy:

0.5*m*v² = m*g*L*sinθ + µ*m*g*L*cosθ

solve for L:

0.5*v² = g*L*(sinθ + µ*cosθ)

L = 0.5*v²/[ g*(sinθ + µ*cosθ)]

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