Question

Figure below show the two block of mass mA = 5 kg and mB = 12 kg suspended by a string on either side of a frictionless pulley. (a) Find the acceleration of the two blocks aA and aB (b) Find the tension of the string (T) (a) aA = 6.91 m/s^2 aB = 6.91 m/s^2 (b) 53.5 N (a) aA = 4 m/s^2 aB = 4 m/s^2 (b) 69 N (a) aA = 0.7 m/s^2 aB = 0.7 m/s^2 (b) 52.5 N (a) aA = 4.2 m/s^2 aB = 4.2 m/s^2 (b) 75 N

          Figure below show the two block of mass mA = 5 kg and mB = 12 kg suspended by a string on either side of a frictionless pulley.
(a) Find the acceleration of the two blocks aA and aB
(b) Find the tension of the string (T)

(a) aA = 6.91 m/s^2 aB = 6.91 m/s^2 (b) 53.5 N
(a) aA = 4 m/s^2 aB = 4 m/s^2 (b) 69 N
(a) aA = 0.7 m/s^2 aB = 0.7 m/s^2 (b) 52.5 N
(a) aA = 4.2 m/s^2 aB = 4.2 m/s^2 (b) 75 N

Figure below show the two block of mass mA = 5 kg and mB = 12 kg suspended by a string on either side of a frictionless pulley.
(a) Find the acceleration of the two blocks aA and aB
(b) Find the tension of the string (T)

(a) aA = 6.91 m/s^2 aB = 6.91 m/s^2 (b) 53.5 N
(a) aA = 4 m/s^2 aB = 4 m/s^2 (b) 69 N
(a) aA = 0.7 m/s^2 aB = 0.7 m/s^2 (b) 52.5 N
(a) aA = 4.2 m/s^2 aB = 4.2 m/s^2 (b) 75 N

Added by Crystal R.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Ajay Singhal

10/24/2021

Step 1

- For block \( A \) (mass \( m_A = 5 \) kg), the force due to gravity is \( F_{gA} = m_A \cdot g = 5 \cdot 9.8 = 49 \) N. - For block \( B \) (mass \( m_B = 12 \) kg), the force due to gravity is \( F_{gB} = m_B \cdot g = 12 \cdot 9.8 = 117.6 \) N. Show more…

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Figure below show the two block of mass mA = 5 kg and mB = 12 kg suspended by a string on either side of a frictionless pulley. (a) Find the acceleration of the two blocks aA and aB (b) Find the tension of the string (T) (a) aA = 6.91 m/s^2 aB = 6.91 m/s^2 (b) 53.5 N (a) aA = 4 m/s^2 aB = 4 m/s^2 (b) 69 N (a) aA = 0.7 m/s^2 aB = 0.7 m/s^2 (b) 52.5 N (a) aA = 4.2 m/s^2 aB = 4.2 m/s^2 (b) 75 N