I always think I know the answer but it never excepts my inputs- I am almost in tears over this homework.1)A truck stops in a distance Dm 254.84 mThe initial velocity is v0 50 m/s and total mass is m 3137.8 kgThe coefficient of friction is mk 0.5 What would the stopping distance have been it a truck had twice the mass as the first? 2) A car with a mass of 500 kg, starting from rest, 402 m drag race in 4.3 sThe final speed is 115 m/sWhat is the engine's average power output?3)A 0.156-kg baseball is thrown, and approaches a bat at a speed of 58.1 m/sThe bat does Wnc 71.6 J of work on the ball in hitting itIgnoring the air resistance, determine the speed of the ball after the ball leaves the bat and is 29.2 m above the point of impact4)A piano with a mass of 191 kg is lifted at a steady speed 14.8m above the groundThe crane that is doing the lifting produces a steady power of 435WHow much time does ii take to lift the piano?
(1) v(f) 0 m/s, set m 6275.60 kg F(friction) uN umg (0.5)(6275.60)(9.8) 30750.44 N F ma or a F/m ug 4.9 m/s? Note here that the acceleration is independent of the mass of the truckv?(f) v?(i) + 2ad d v?(i)/2a Hence, doubling the mass of the truck doesn't change the stopping distance(2) AVG Power Total energy/Total time W Fd mad m(Δv/Δt)d (500) [(115)/(4.3)] (402) 5,375,581.39 joules AVG Power W/t 1,250,135.21 watts or about 1,675 horsepower (3) Here you will need to split the problem into two parts First calculate the energy of the ball after being struck by the bat: E(total) W(bat) + ?mv?(i) (71.6) + (0.5)(0.156)(58.1)? 334.90 joules Now the ball is under the influence of gravity E(total) mgh + ?mv?(f) v(f) sqrt[2(E(total) - mgh)/m] 61 m/s (4) PE W mgh Power W/t mgh/t t (191)(9.8)(14.8)/(435) 63.68 sec 5) An object of mass 4.3 kg is projected into the air at a 50° angleIt hits the ground 3.4 s laterWhat is its change in momentum while it is in the air? Ignore air resistanceΔp mΔv mgΔt (4.3)(9.8)(3.4) 143.28 kg m/s