Respuesta :
Answer:
-4892 J
Explanation:
The work done by friction is equal to the difference in mechanical energy of the car between the bottom of the loop (point A) and the top of the loop (point B).
The mechanical energy of the car at point A is
[tex]E_A = \frac{1}{2}mv_A^2 + mgh_A[/tex]
where
[tex]m=108 kg[/tex] is the mass of the car
[tex]v_A = 25.0 m/s[/tex] is the speed at point A
[tex]h_A=0[/tex] is the height of the car at point A (zero because it is at the bottom of the loop)
Substituting into the equation, we find
[tex]E_A = \frac{1}{2}(108 kg)(25.0 m/s)^2 + (108 kg)(9.8 m/s^2)(0)=33,750J[/tex]
The mechanical energy of the car at point B is
[tex]E_B = \frac{1}{2}mv_B^2 + mgh_B[/tex]
where
[tex]m=108 kg[/tex] is the mass of the car
[tex]v_B = 8.0 m/s[/tex] is the speed at point B
[tex]h_B=24.0 m[/tex] (twice the radius) is the height of the car at point B, at the top of the loop
Substituting into the equation, we find
[tex]E_B = \frac{1}{2}(108 kg)(8.0 m/s)^2 + (108 kg)(9.8 m/s^2)(24.0 m)=28,858J[/tex]
So, the work done by friction is
[tex]W=E_B-E_A=28,858 J-33,750 J=-4,892 J[/tex]
And the work is negative because it is done against the motion of the car.
The work done by friction on the roller coaster is 30,294 J.
Work energy theorem
The work done by friction on the roller coaster is the change in the kinetic energy of the roller coaster during the journey from bottom to top of the loop.
W = ΔK.E
W = ¹/₂mv₂² - ¹/₂mv₁²
W = ¹/₂m(v₂² - v₁²)
W = ¹/₂ x 108(8² - 25²)
W = -30,294 J
Thus, the work done by friction on the roller coaster is 30,294 J.
Learn more about work done by friction here: https://brainly.com/question/12878672