During a rockslide, a 570 kg rock slides from rest down a hillside that is 700 m along the slope and 210 m high. the coefficient of kinetic friction between the rock and the hill surface is 0.13. (a) if the gravitational potential energy u of the rock-earth system is zero at the bottom of the hill, what is the value of u just before the slide? (b) how much energy is transferred to thermal energy during the slide? (c) what is the kinetic energy of the rock as it reaches the bottom of the hill? (d) what is its speed then?

Answered

Explanation:

(a)

        Gravitational potential energy :

                  U =   mgh = 570 kg x 9.8 x 210 = 1.173 x 10^6   J

(b)

         We know that f_K = μ_K ×N     where N = Normal reaction = mg cos θ

                                     = μ_K mg cosθ

From the  diagram in the attachment :   cosθ = BC / AB = 70√91 /  700

         ∴  f_K   = 0.13 x 5700 x 9.8 x 70√91 / 700

Thermal energy during the slide energy :

                   E =   f_K x l

                       =  ( 0.13 x 570 x 9.8 x 70√91 / 700) x 700

                        = 4.84 x 10^5 J

(c)

            we know that :

                     Δ E = - ( Δ K + ΔU )

                  ∴ Δ K =   U_i - U_f - ΔE

                                = 1.173 x 10^6 - 0 - 4.84 x 10^5   = 6.89 x 10^5 J

(d)

       From the above

                K = 6.89×10^5   J

            1/2 m v^2   =  6.89×10^5

                  from the above equation solving for v we get

              v = 49.20 m/s

Hope this helps u!

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