Two blocks connected by a string are on a horizontal frictionless surface. The b

Question

Two blocks connected by a string are on a horizontal frictionless surface. The blocks are connected to a hanging weight by means of a string that passes over a pulley as shown in the figure below, where m1 = 1.75 kg, m2 = 2.80 kg, and m3 = 4.90 kg.

(a) Find the tension T in the string connecting the two blocks on the horizontal surface. ( Answer given is 8.89 N... HOW DO I GET THIS ANSWER)

(b) How much time is required for the hanging weight to fall 10.0 cm if it starts from rest? ( Answer given is .198s ... HOW DO I GET THIS ANSWER)

Explanation / Answer

Consider all 3 masses as 1 system initially because they will all slide together with the equal acceleration. The net force on the system is the weight of the m3 which is 48.02 N (4.90 kg x 9.8 m/s2). There isn't air resistance force since the surface is frictionless. The total mass of the system is 9.45 kg. Use Newton's 2nd law youcan calculate the acceleration of the system. 48.02 N/ 9.45 kg= 5.08 m/s2. This is the acceleration. a) Now, you can use the accleration to find the tension(T1). Look at the block m1. The only force acting on m1 is thetension, so this is the net force. You have the acceleration (5.08 m/s2), and the mass (1.75 kg), so using Newton's 2nd law again, solve for the Force. F= 1.75 kg x 5.08 m/s2= 8.89 N. b) Use 5.08 m/s2 as the acceleration. Initial velocity is 0m/s.  x is .1 m (i.e. 10cm) . Using the equation of motion, x= vi t + 1/2at2, you may now solve to get t. vi = 0 so x =1/2at2    or    t = 2x/a  = sqrt(2*0.1/5.08)=0.198 s Consider all 3 masses as 1 system initially because they will all slide together with the equal acceleration. The net force on the system is the weight of the m3 which is 48.02 N (4.90 kg x 9.8 m/s2). There isn't air resistance force since the surface is frictionless. The total mass of the system is 9.45 kg. Use Newton's 2nd law youcan calculate the acceleration of the system. 48.02 N/ 9.45 kg= 5.08 m/s2. This is the acceleration. a) Now, you can use the accleration to find the tension(T1). Look at the block m1. The only force acting on m1 is thetension, so this is the net force. You have the acceleration (5.08 m/s2), and the mass (1.75 kg), so using Newton's 2nd law again, solve for the Force. F= 1.75 kg x 5.08 m/s2= 8.89 N. b) Use 5.08 m/s2 as the acceleration. Initial velocity is 0m/s.  x is .1 m (i.e. 10cm) . Using the equation of motion, x= vi t + 1/2at2, you may now solve to get t. vi = 0 so x =1/2at2    or    t = 2x/a  = sqrt(2*0.1/5.08)=0.198 s

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