A crane lifts a 425 kg steel beam vertically a distance of 117m. How much work does the crane do on the beam if the beam accelerates upward at 1.8 m/s^2? Neglect frictional forces.a.) 3.4 x 10^5 Jb.) 4.0 x 10^5 Jc.) 5.8 x 10^5 Jd.) 4.9 x 10^5 JSolved as W= mgh but obviously that is wrong. Where do I go wrong?
You have the right idea, the answer is d. EDIT: Well I actually read the rest of the question (sometimes I get an itchy trigger finger!), and the net force should be found first, the net force includes the weight and the tension in the crane's cable (so you need to find the tension): ∑F = ma = T - mg T = m(a + g) = 425kg(1.80m/s? + 9.80m/s?) = 4930N So the net force is the sum of the tension in the cable and the weight of the beam, or: F = T - mg = 4930N - (425kg)(9.80m/s?) = 735N Then, the work done by the crane is: W = FΔy = 735N(117m) = 8.95 x 10^4J Does not match any of your choices, so I am perplexed! Maybe someone will straighten me out. Apologies if I failed to help.
Force=mass x (acceleration) -Gravity must be taken into account F=425 x (1.8 -(-9.8)) F=4930 w= F x deltay w= 4930 x 117 w=576810J
