A 1400-car is accelerating up a hill. The hill is 150 M long and the total rise of the hill is 6.0 M. The car accelerates from a speed of 7.0 m/s at the bottom to 15 m/s at the top. If the average retarding force of friction is 700 N, find the change in potential energy of the car, and the change in the kinetic energy of the car.
Show the work!!!!
Show the work!!!!
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At the bottom:
PE=0; KE=0.5mv^2=0.5*1400*7^2 J = 34300 J
At the top:
PE=mgh=1400*9.81*6 J = 82404 J
KE=0.5mv^2=0.5*1400*15^2 = 157500 J
Total energy = 239904 J
Change in energy = 239904 - 34300 = 205604 J
PE=0; KE=0.5mv^2=0.5*1400*7^2 J = 34300 J
At the top:
PE=mgh=1400*9.81*6 J = 82404 J
KE=0.5mv^2=0.5*1400*15^2 = 157500 J
Total energy = 239904 J
Change in energy = 239904 - 34300 = 205604 J
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Initial potential energy of car P1 =0 (at ground level)
Final potential energy of car P2= mgh= 1400*9.8*6 Nm
change in P.E = P2-P1
Initial Kinetic energy of car K1= mu^2/2 Final Kinetic Energy K2= mv^2/2
Change in Kinetic energy K2-K1= 0.5*1400*(15^2 -7^2)
Final potential energy of car P2= mgh= 1400*9.8*6 Nm
change in P.E = P2-P1
Initial Kinetic energy of car K1= mu^2/2 Final Kinetic Energy K2= mv^2/2
Change in Kinetic energy K2-K1= 0.5*1400*(15^2 -7^2)