In a setup like that shown in Fig. 6-37, a 4.34 kg block starts from rest on the

Question

In a setup like that shown in Fig. 6-37, a 4.34 kg block starts from rest on the left edge of a frictionless tabletop 1.30 m wide. It accelerates to the right, and reaches the right edge in 2.84 s. If the mass of the block hanging from the left side is 3.26 kg, what is the mass hanging from the right side?

Explanation / Answer

The acceleration of the 4.34-kg block sliding horizontally to the right across the frictionless tabletop is a = 2(1.3m)/(2.84)2=0.322 m/s2 This is also the magnitude of the accelerations of the other two blocks. Since we are not interested in the tension in either string, we may use a shortcut to find the unknown mass. The net force on all three masses in the direction of motion is the difference in the right- and left-hand weights, Fnet = (m ? 3.26kg)g , which equals the total mass times the acceleration, (m + 3.26kg + 4.34kg)a . Solving for m, we find m = [(3.26 kg + 4.34 kg )(0.322 m/s 2 ) + (3.26 kg )(9.81 m/s 2 )]/(9.81 m/s 2 ? 0.322 m/s 2 ) = 3.628 kg ---> Answer (The shortcut can be justified by adding the component of the equations of motion of the three blocks in the direction of motion: Tleft ? (3.26kg)g = (3.26kg)a , Tright ? Tleft = ( 4.34kg)a , and mg ? Tright = ma . In this setup, the ropes are massless, and the pulleys are massless and frictionless, so the tension in each rope is constant.)

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