A block of mass m begins at rest at the top of a ramp at elevation h with whatev

Question

A block of mass m begins at rest at the top of a ramp at elevation h with whatever PE is associated with that height.

The block slides down the ramp over a distance d until it reaches the bottom of the ramp. How much of its original total energy (in J) survives as KE when it reaches the ground?

(In other words, the acceleration is not zero like it was in lab and friction does not remove 100% of the original PE. How much of that original energy is left over after the friction does work to remove some?)

m = 5.0 kg
h = 8.5 m
d = 5 m
= 0.3
= 36.87°

Explanation / Answer

Given

  
m = 5.0 kg

h = 8.5 m

d = 5 m

= 0.3

= 36.87°

the initial energy of the system is E = k.e+p.e = 0+mgh = 5*9.8*8.5 J = 416.5 J

now

as the blok moves down the ramp the fricitonal force also acts on it so the forces acting ar e

  
   F = mg sin theta- muek*mg cos theta
   = mg(sin theta - muek cos theta)
   = 5*9.8 (sin 36.87 - 0.3cos36.87) N

   = 17.640 N

work done = change in kinetic energy

   F*d = 0.5*m(v2^-v1^2)

   = 0.5m(v2^2 -0)

   v2^2 = 2*F*d/m

   = 2*17.640*5/5

   = 35.28

   v2 = sqrt(35.28) = 5.94 m/s

the energy at tthe bottom is E = p.e+k.e = 0 +0.5*5*5.94^2 = 88.209J

initial energy at top = 416.5 J , at bottom kinetic energy is 88.209 J

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