A crane lifts a 2355 N bucket containing 1.81 m3 of soil (density=2084 kg/m3) to a height of 7.56 m. Calculate the work the crane performs.
W=Fdcos(angle) W=mgh=2355(7.56)=17803.8 J
Mass of soil(m) = V x d =m = 1.81 x 2084 = 3772.04 kg =Weight = m x g = 3772.04 x 9.8 = 36965.99 N Thus the total Force by crane = 36965.99 + 2355 = 39320.99 N Thus Work done (W) = F x s = 39320.99 x 7.56 = 297266.70 J = 297.27 kJ
I will try to make it very simple. First lets convert the 2355 into a mass. 2355/9.81 = 240.06 Kg Then, we need to get the mass of the soil. Density (p)= m/v m= pv m= 2084*1.81 = 3772.04 Kg Total mass= 4012.1 Kg The final step is to calculate the work done. It can be solved from two perspectives and both will lead to the same solution. Sol. 1: Since that the crane is lifting an object, then the work the crane is doing is against gravity which is calculating the gravitational potential energy gained by the bucket. GPE= mgh= 4012.1*9.81*7.56 = 2.99*10^5 Sol 2: Work done= Force*distance moved in the direction of the force The force in this case will be the weight of the bucket. Work done= (4012.1*9.81)*7.56 = 2.99*10^5
