Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

A girl in a wheelchair (total mass 50kg) is traveling at 1.2 m/s at the crest of

ID: 1517693 • Letter: A

Question

A girl in a wheelchair (total mass 50kg) is traveling at 1.2 m/s at the crest of a slope 2.7m high and 11m long. At the bottom of the slope her speed is 6.5 m/s. All drag and friction forces can be modeled as a constant backward force equal to 47N. Find the work she did pushing forward on the wheels. A girl in a wheelchair (total mass 50kg) is traveling at 1.2 m/s at the crest of a slope 2.7m high and 11m long. At the bottom of the slope her speed is 6.5 m/s. All drag and friction forces can be modeled as a constant backward force equal to 47N. Find the work she did pushing forward on the wheels. A girl in a wheelchair (total mass 50kg) is traveling at 1.2 m/s at the crest of a slope 2.7m high and 11m long. At the bottom of the slope her speed is 6.5 m/s. All drag and friction forces can be modeled as a constant backward force equal to 47N. Find the work she did pushing forward on the wheels.

Explanation / Answer

Given

   mass m = 50 kg, speed at crest v1 = 1.2 m/s, and at bottom of the slope is V2 = 6.5 m/s

   height h = 2.1 m, length l= 11 m, F = 47 N


if we write the conservatio of energy at both positions

   toata energy of the system is constant

  
U_g + k.e_1 = k.e_2 + E_t +U_o

   U_g - gravitational potential energy
k.e_1-kinetic energy at crest (top of the ramp)
   k.e_2 - kinetic energy at bottom point   
   E_t - thermal energy due to friction
   U_o - work done by the girl in down the hill ride

mgh + 1/2 mv1^2 = 1/2 mv2^2 + F*X+ U_O
  
   50*9.8*2.1+ 0.5*50*1.2^2 = 0.5*50*6.5^2+ 47*11+U_O

       U_o= - 508.25 J

work done by the girl pushing forward on the wheels downhill ride is 508.25 J

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote