Car rolling down a 55.0m hill. Changes elevation by 1.00m every 4.00m traveled. What is velocity at bottom of hill if coefficient of friction is 0.10?

Question

Significant Figures Apply

Detailed Calculations

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Simple Minded Man

Answer 1

This problem can be solved using the principle that external work done on a system is equal to the change in energy of the system.   In symbols

 

    W_ext =  ΔE  =  E_f - E_i     (here E is KE +PE)

 

I am going to assume that 55.0 m hill means the the slant length of the hill is 55.0 m.   This implies that vertical drop is 55.0 /4 =  13.75m.  

 

The external work is the work done by friction.  The force of friction is   m g μ cos(θ)  .  Here μ = 0.10 and cos(θ) = cos(arctan(1/4) ) = 0.970.   The work done by friction is the negative of this force times the distance traveled (55.0m)  .

 

The change in energy is  the kinetic energy at the bottom of the hill  minus the drop in elevation (13.75) times mg .  In symbols the equation becomes

 

    - m g μ cos(θ) 55.0 =  (1/2) m  v²  - m g (55.0/4)

 

Notice the factors of m all cancel out.   This cancellation leaves

 

  v²  =  2[ (9.8) (55.0/4) - (9.8) (0.10) (0.970) 55.0]   =  165.0  

and    v =  12.8 m/s

 

The factor of cos(arctan(1/4)) appears because the normal force for an inclined plane geometry is the weight times the cosine of the inclined plane angle, θ.  Since the grade is 1/4   , θ = arcgtan(1/4).   The force of friction is the normal force times the coefficient of friction, μ .