Rock Slide During a rockslide, a 330 kg rock slides from rest down a hillside th

Question

Rock Slide During a rockslide, a 330 kg rock slides from rest down a hillside that is500 m long and 300 m high. The coefficient of kinetic frictionbetween the rock and the hill surface is 0.34. (a) If the gravitational potential energyU of the rock-Earth system is set to zero at the bottom ofthe hill, what is the value of U just before theslide?
1 J
(b) How much energy is transferred to thermal energy during theslide?
2 J
(c) What is the kinetic energy of the rock as it reaches the bottomof the hill?
3 J
(d) What is its speed then?
4 m/s

Section 10.7 Conservation of Energy (a) If the gravitational potential energyU of the rock-Earth system is set to zero at the bottom ofthe hill, what is the value of U just before theslide?
1 J
(b) How much energy is transferred to thermal energy during theslide?
2 J
(c) What is the kinetic energy of the rock as it reaches the bottomof the hill?
3 J
(d) What is its speed then?
4 m/s

Explanation / Answer

           Given that the mass of the rock is M = 330 kg            height h = 300 m and lenght L = 500 m            The coefficient of kinetic friction between the rock and the hillsurface is = 0.34 -----------------------------------------------------------------------------------------------     From the given data the angle ofinclination is = sin-1( 300m /500m) =36.860      (a) The potential energy is zero atthe bottom then                 Total energy just before reach the bottom is K = Mgh - mgcos*L                                                                                         =--------- J     (b)   The energytransfered as thermal energy is E = mgcos*L                                                                                 =----------- J          This is onlydue to friction .    (c) The kinetic energy ofthe rock just beforce reach the bottom is                    K= Mgh - mg cos*L                        =--------- J    (d ) Then the speed is (1/2)mV2=   Mgh - mg cos*L                                              then we get V = ----------- m/s                     -----------------------------------------------------------------------------------------------     From the given data the angle ofinclination is = sin-1( 300m /500m) =36.860      (a) The potential energy is zero atthe bottom then                 Total energy just before reach the bottom is K = Mgh - mgcos*L                                                                                         =--------- J     (b)   The energytransfered as thermal energy is E = mgcos*L                                                                                 =----------- J          This is onlydue to friction .    (c) The kinetic energy ofthe rock just beforce reach the bottom is                    K= Mgh - mg cos*L                        =--------- J    (d ) Then the speed is (1/2)mV2=   Mgh - mg cos*L                                              then we get V = ----------- m/s                                            =--------- J    (d ) Then the speed is (1/2)mV2=   Mgh - mg cos*L                                              then we get V = ----------- m/s

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