Question

Two blocks of masses 5 kg and 7 kg are connected by uniform rope of mass 4 kg as shown in the figure.
An upward force F = 200 N is applied on the system. (Take g = 10 m/s2)

4. The net downward force of the system is
(A) 40 N (B) 160 N
(C) 200 N (D) 16 N
5. The acceleration of the system is
(A) 2.5 m/s2 (B) 5 m/s2
(C) 6 m/s2 (D) 7.5 m/s2
6. The tension at the top of the heavy rope is
(A) 100 N (B) 112.5 N
(C) 125 N (D) 150 N
7. The tension at the midpoint of the rope is
(A) 50 N (B) 75 N
(C) 87.5 N (D) 98.5 N
8. The tension at the lower end point of the rope is
(A) 50 N (B) 75 N
(C) 87.5 N (D) 62.5 N

Answer 1

Given,

Mass of top block (m1) = 5 kg

Mass of lower block (m2) = 7 kg

Mass of rope (mw) = 4 kg

The total mass of the system (m) = 5+ 4+ 7

                                                       16 kg

Force = 200 N

4) total downward force  = mg

                                         = 16 × 10

                                         = 160 N

thus, the correct option is B

5) the accereleration of the system,

According to the equation

F=ma

=> F - mg = ma (g= 10 m/s^2

=> 200 - 16 × 10 = 16 a

=> 200 - 160 = 16 a

=> 40/16 = a

=> 2.5 = a

∴ acceleration of the system = 2.5 m/s^2

6) The tension at the top of the heavy rope is of mass

mass = 5+ 4 = 9 kg

a = 2.5 m/s^2

T= ma + mg

  =m (a+ g)

  = 9 × 12.5

  = 112.5 N

7) Tension at the midpoint of the rope,

m = 4 kg

T = m(a +g)

   = 4 × 12.5

   = 50 N

8) Tension at the lower end point of the rope

   m = 7kg

T = m (a+g)

  = 7 × 12.5

   = 87.5 N

Answer 2

Answer:

Explanation:

Given,

Mass of top block (m1) = 5 kg

Mass of lower block (m2) = 7 kg

Mass of rope (mw) = 4 kg

The total mass of the system (m) = 5+ 4+ 7

                                                      16 kg

Force = 200 N

4) total downward force  = mg

                                        = 16 × 10

                                        = 160 N

thus, the correct option is B

5) the accereleration of the system,

According to the equation

F=ma

=> F - mg = ma (g= 10 m/s^2

=> 200 - 16 × 10 = 16 a

=> 200 - 160 = 16 a

=> 40/16 = a

=> 2.5 = a

∴ acceleration of the system = 2.5 m/s^2

6) The tension at the top of the heavy rope is of mass

mass = 5+ 4 = 9 kg

a = 2.5 m/s^2

T= ma + mg

 =m (a+ g)

 = 9 × 12.5

 = 112.5 N

7) Tension at the midpoint of the rope,

m = 4 kg

T = m(a +g)

  = 4 × 12.5

  = 50 N

8) Tension at the lower end point of the rope

  m = 7kg

T = m (a+g)

 = 7 × 12.5

  = 87.5 N

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