Three blocks of masses m1, m2, and m3 are placed on a horizontal frictionless surface. A force of 40 N pulls the system. Calculate the value of T, if:
Correct answer: B. 20 N
- A. 40 N
- B. 20 N
- C. 10 N
- D. 5 N
Explanation
The following is the solution:Let a be the acceleration of the system,F is the net pulling force on m1,T is the tension force exerted by m1 on m2 and by m2 on m1,A different tension force T' is exerted by m2 on m3, and by m3 on m2, as shown in the diagram. (While this force exists, we're not interested in it because the question doesn't ask for it). The blocks move as a single system, therefore, by Newton's second law:a = F/(m1+m2+m3)a = 40 / (10+6+4) = 40/20a = 2 m/s²Now, we need to find the tension force T acting on m1 in the direction opposite to the 40 N force. Consider the free-body diagram for m1. F - T = m1 × a40 - T = 10 × 2T = 20 N
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