A block of mass m1 = 42.3 kg on a horizontal surface is connected to a mass m2 =

Question

A block of mass m1 = 42.3 kg on a horizontal surface is connected to a mass m2 = 18.0 kg that hangs vertically as shown in the figure below. The two blocks are connected by a string of negligible mass passing over a frictionless pulley. Assume that the horizontal surface is smooth.
(a) What is the acceleration of the hanging mass? (Enter the magnitude only.) m/s2
(b) Determine the tension in the cord. (Enter the magnitude only.) N A block of mass m1 = 42.3 kg on a horizontal surface is connected to a mass m2 = 18.0 kg that hangs vertically as shown in the figure below. The two blocks are connected by a string of negligible mass passing over a frictionless pulley. Assume that the horizontal surface is smooth.
(a) What is the acceleration of the hanging mass? (Enter the magnitude only.) m/s2
(b) Determine the tension in the cord. (Enter the magnitude only.) N A block of mass m1 = 42.3 kg on a horizontal surface is connected to a mass m2 = 18.0 kg that hangs vertically as shown in the figure below. The two blocks are connected by a string of negligible mass passing over a frictionless pulley. Assume that the horizontal surface is smooth.
(a) What is the acceleration of the hanging mass? (Enter the magnitude only.) m/s2
(b) Determine the tension in the cord. (Enter the magnitude only.) N

Explanation / Answer

let acceleration be a

the forces acting on m1 are : tension of string = T rightward

forces acting on m2 are : tension of string =T upward and grvitational force = m2*g downward

since same string therefore tension will be same

now the equations are : m1*a = T .................(1)

m2*a= m2*g - T ......(2)

putting value of T from 1 into eq (2) we get , m2*a=m2*g-m1*a

substituting values : 18*a=18*9.8 - 42.3 *a

60.3*a = 18*9.8

=> a = 2.92 m/s^2

putting in (1) we get, T= 42.3*2.92 = 123.51 N ans

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