In Fig. E6.7 assume that there is no friction force on the 20.0-N block that sits on the tabletop. The pulley is light and frictionless. (a) Calculate the tension Tin the light string that connects the blocks. (b) For a displacement in which the 12.0-N block descends 1.20 m, calculate the total work done on (i) the 20.0-N block and (ii) the 12.0-N block. (c) For the displacement in part (b), calculate the total work done on the system of the two blocks. How does your answer compare to the work done on the 12.0-N block by gravity? (d) If the system is released from rest, what is the speed of the 12.0-N block when it has descended 1.20 m?
Solution 17E Step 1: (a).First we have to find masses to calculate tension.the tension formula is given by T = [(m1 * m2)/(m 1 m )2 g* m1= 20 N/9.8 m/s = 2.04 kg m2= 12 N/9.8 m/s = 1.22 kg 2 T = [(2.04 kg *.22 kg)/(2.04 kg + 1.22 kg)] *.8 m/s 2 T = [(0.76319)] *.8 m/s T = 7.4792~7.5 N.