The kinetic energy halfway the hill is [tex]2.86\cdot 10^5 J[/tex]
Explanation:
If there are no friction forces acting on the cart, we can apply the law of conservation of energy: the mechanical energy of the cart (which is the sum of potential energy + kinetic energy) must be conserved. So we can write:
[tex]U_A +K_A = U_B + K_B[/tex]
where
[tex]U_A=mgh_A[/tex] is the initial potential energy, at point A, with
m = 980 kg (mass of the cart)
[tex]g=9.8 m/s^2[/tex] (acceleration of gravity)
[tex]h_A = 30 m[/tex] (height at point A)
[tex]K_A=\frac{1}{2}mv_A^2[/tex] is the initial kinetic energy, at point A
, with
[tex]v_A=17 m/s[/tex] (velocity at point A)
[tex]U_B=mgh_B[/tex] is the final potential energy, at point B, where
[tex]h_B = 15 m[/tex] (height at point B)
[tex]K_B=\frac{1}{2}mv_B^2[/tex] is the final kinetic energy, at point B, where
[tex]v_B[/tex] is the velocity at point B
Here we are interested in finding [tex]K_B[/tex], so by re-arranging the equation and substituting we find:
[tex]K_B = U_A+K_B-U_B = mg(h_A-h_B)+\frac{1}{2}mv_A^2=(980)(9.8)(30-15)+\frac{1}{2}(980)(17)^2=2.86\cdot 10^5 J[/tex]
Learn more about kinetic energy:
brainly.com/question/6536722
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