Recent content by Kira

err okay a friend told me I am not supposed to put it into w = fd because by solving for the intergral we get work since force is a factor or distance...is the person correct?

so its F = 15000X + 5000X² - (25000/3)x³ F(.6) = 15000(.6) + 5000(.6²) - (25000/3)(.6³) F = 9000N W = fd w = 9000(.6) w = 5400 J

wait so w = Integral(F) ? that seems a little more farfetched... since that's still a pretty big equation..

okay for number 2 F(x) = 15000 + 10000x - 25000x² F'(x) = 10000 - 50000x F'(.6) = 10000 - 50000(.6) F'(6) = -20000 w = fd w = -(20000)(.6) w = -12000 J err I don't think you can have negative Joules...

Basically my teacher hands us out a review packet for physics and says to do it if we want practice...Okay I do the packet and I have it done, now my problem is that the teacher didn't give out the answers to the problems so I have no idea whether or not I did it right...so I am hoping some one...

okay thanks for the reply

okay I did a little more work but I don't know if the concepts are correct a. Force normal (from the road) would be equal to weight minus centripetal force F = mg - mv²/r F = (1800)(9.8) - (1800)(16²)/42 F = 6668.6 N b.weight = centripetal force mg = mv²/r v = sqrt(mgr/m) v...

An 1800 kg car passes over a hump in a road that follows the arc of a circle of radius 42 m. a. What force does the road exert on the car as the car passes the highest point of the hump if the car travels at 16 m/s? b. What is the maximum speed the car can have as it passes this highest...