10. [41 Points] SERCP104.AE.014. EXAMPLE 4.14 Connected Objects GOAL Use both the general method and the system -D 11 approach to solve a connected two-body problem involving "'1 l gravity and friction. 4'13) ml _) '32 PROBLEM (a) A block with mass m1 = 4.00 kg and a T I ball with mass m2 = 7.00 kg are connected by a light string that passes over a frictionless pulley, as shown in MI? figure (a). The coefficient of kinetic friction between the block and the surface is 0.300. Find the acceleration of the a two objects and the tension in the string. (b) Check the answer for the acceleration by using the system approach. '-]L T" 2 STRATEGY Connected objects are handled by applying L Newton's second law separately to each object. The force diagrams for the block and the ball are shown in figure (b), with the +x-direction to the right and the +y-direction LI} upwards. The magnitude of the acceleration for both (a) Two objects connected by a lightming that passes objects has the same value Ia I = la I = a_ The bIOCk overa frictionless pulley. (b) Force diagrams for the ' 1 2 objects. with mass m1 moves in the positive x-direction, and the ball with mass m2 moves in the negative y- I _ ) V J)! 2g direction, so a1 = -a2. Using Newton's second law, we can develop two equations involving the unknowns T and a that can be solved simultaneously. In part (b), treat the two masses as a single object, with the gravity force on the ball increasing the combined object's speed and the friction force on the block retarding it. The tension forces then become internal and don't appear in the second law. SOLUTION (A) Find the acceleration of the objects and the tension in the string. Write the components of Newton's ZFX = T fk = \"\"131 Z}: = n _ mtg = 0 second law for the block of mass m 1' y The equation for the y-component (1) T Jukmlg = mla1 gives :2 = mlg. Substitute this value for n and fl: = ,ukn into the equation for the x-component. Apply Newton's second law to the ball, Fy = T - m29 = maz = -m2a1 recalling that a 2 = - a 1. ( 2 ) T - m29 = -m2a1 Subtract Equation (2) from Equation m2g - ukm19 = (m1 + m2)a1 (1), eliminating T and leaving an m29 - Him19 equation that can be solved for a 1. a1 = - my + m 2 Substitute the given values to obtain a1 = (7.00 kg) (9.80 m/s2) - (0.300) (4.00 kg) (9.80 m/s2) the acceleration. (4.00 kg + 7.00 kg) = 5.17 m/s2 Substitute the value for a, into T = 32.4 N Equation (1) to find the tension T. (B) Find the acceleration using the system approach, where the system consists of the two blocks. Apply Newton's second law to the (m1 + m2)a = m29 - ukn = m29 - ukm19 system and solve for a. m2g - ukm 19 a my + m2 LEARN MORE REMARKS Although the system approach appears quick and easy, it can be applied only in special cases and can't give any information about the internal forces, such as the tension. To find the tension, you must consider the free-body diagram of one of the blocks separately as was done in part (a) of this example. QUESTION If mass m, is increased, does the acceleration of the system increase, decrease, or remain the same? O increases O decreases O remains the same Does the tension increase, decrease, or remain the same? O increases O decreases O remains the same PRACTICE IT Use the worked example above to help you solve this problem. A block with mass m, = 4.50 kg and a ball with mass m, = 7.30 kg are connected by a light string that passes over a frictionless pulley, as shown inPRACTICE IT Use the worked example above to help you solve this problem. A block with mass m1 = 4.50 kg and a ball with mass m2 = 7.30 kg are connected by a light string that passes over a frictionless pulley, as shown in figure (a). The coefficient of kinetic friction between the block and the surface is 0.300. (3) Find the acceleration of the two objects and the tension in the string. a =Ews2 E T: N (b) Check the answer for the acceleration by using the system approach. (Use the following as necessary: m1, m2, pk, and g.) EXERCISE HINTS: GETTING STARTED | I'M STUCK! Use the values from PRACTICE IT to help you work this exercise. What if an additional mass is attached to the ball? How large must this mass be to increase the downward acceleration by 40%? Em Why isn't it possible to add enough mass to double the acceleration? This answer has not been graded yet