The block slides at a constant speed on a surface with friction. In each case, the water jet has velocity 50 m/s and cross-sectional area 5 cm2, and theblock has velocity 25 m/s. In case (a), the jet is deflected to an angle of 60° from horizontal relative to the block, as shown. In case (b), the jet isdeflected 180°. Calculate the friction force in each case. Apply both a moving Reference Frame (RF) and a fixed RF. Use the VSM. Check that both RFsgive the same friction force for a given case. (We would not expect the two cases to have the same friction force!) Density of water is 1000 kg/m3. Alljets are open to the atmosphere.case (a)60°case (b)

Given:- The velocity of the water jet = 50 m/s The cross-sectional area = 5 cm2 The velocity of the block = 25 m/s For case (a), the jet is deflected at an angle = 60 degree For case (b), the jet is deflected at an angle = 180 degree To determine:- The frictional force in both caseThe density of the water = 1000 kg/m3 For case (a), The force required to move the block will be Force = ρ×Q×∆VForce = ρ×A×V-Vb×V-Vb×cos 60° i^+V-Vb×sin 60° j^-V-Vbi^ Here, A = Cross-sectional area V = Velocity of jet = 50 m/s Vb = Velocity of block = 25 m/s Force = ρ×A×50-25×50-25×cos 60° i^+50-25×sin 60° j^-50-25i^Force = ρ×A×25×25×cos 60° i^+25×sin 60° j^-25i^Force = ρ×A×25×252 i^+25×32 j^-25i^Force = ρ×A×25×-252 i^+25×32 j^ Now, The net force in x-direction will be FnetX=0FFrictional+ρ×A×25×-252=0Putting, ρ = 1000 kg/m3, A=5 cm2 = 5104m2FFrictional-1000×5104×6252=0FFrictional=156.25 N Hence, The frictional force in case (a) = 156.25 NFor case (b), The force required to move the block will be Force = ρ×Q×∆VForce = ρ×A×V-Vb×2×V-VbForce = 2×ρ×A×V-Vb2 Putting, A = 5 cm2 = 0.0005 m2 V = Velocity of jet = 50 m/s Vb = Velocity of block = 25 m/s So, The frictional force will be Ffrictional=2×1000×0.0005×50-252Ffrictional=1×252Ffrictional=1×625Ffrictional=625 N Hence, The frictional force in case (b) = 625 N The value of frictional force is not similar in both case