Question

A 9.00-kg hanging weight is connected by a string over a pulley to a 5.00-kg block that is sliding on a flat table Fig. If the coefficient of kinetic friction is 0.200, find the tension in the string.

(A)39. 2A (B)38. 57N (C)37.23N  (D)34.55N (E)35. 66N 

Answer 1

Answer 2

writing the force equation we have

9g - T = 9a --------(1)

T - 0.2*5*g = 5a -----(2)

adding both equations we have

8g = 14a

a = 5.6 m/s2

T = 9(g - a) = 9 (9.8 - 5.6) = 37.8 N

so ooption C) 37.23 N is the closest and is the correct answer

Answer 3

SOLUTION :

Let block 9 kg is moving down at an acceleration of a m/s^2.

So, for 9 g block’s free diagram :

           T

           ^

          _I_

         I __I          =   I    9 a

            I                 V

           V

          9g 

Resultant force on the 9 kg block = T - 9 g 

So, 9 a = 9g - T

=> a = g  - T/9 

Block 5kg moves to the right with the same acceleration I.e.  g - T/9 

The forces acting are :

                            ^  5 g

                           _I_

     5 g µ     < _ _I__ I——> T       =  5 a 

                         //////////   I

                             I   

                            V

                            5 g

 

So,

T - 5g *0.200 = 5 a = 5 (g - T/9)

=> T - g = 5g - 5/9 T

=> T + 5/9 T = 5g  + g = 6g = 6 * 9.8 = 58.8

=> T = 58.8 * 9/14 = 37.8 N 

Hence, Tension in the string is 37.8 N : Option  E is nearest (ANSWER).