Laws of Motion
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Question : 15
Total: 40
Two bodies of masses 10 kg and 20 kg respectively kept on a smooth, horizontal surface are tied to the ends of a light string. A horizontal force F = 600 N is applied to (i) A, (ii) B along the direction of string. What is the tension in the string in each case?
Solution:
Here, F = 600 N ;
m 1 = 10 k g ; m 2 = 20
Let T be tension in the string and a be the acceleration of the system, in the direction of force applied.
(a) When force is applied on lighter block A, as shown in figure (a)m 1 and m 2
F – T = m 1 a
T = m 2 a
Divide (i) by (ii), we get
=
;
− 1 =
=
=
⇒
=
+ 1 =
;
T =
F
=
× 600 = 400 N
From equation (ii),a =
=
= 20 m s − 2
(b) When force is applied on heavier block B as shown in figure (c)T = m 1 a
F – T = m 2 a
Divide (iv) by (iii), we get
=
=
= 2 ;
− 1 = 2 ;
= 3 ,
T =
=
= 200 N
From equation (iii)a =
=
= 20 m s − 2
Let T be tension in the string and a be the acceleration of the system, in the direction of force applied.
(a) When force is applied on lighter block A, as shown in figure (a)
The equation of motion of masses
Divide (i) by (ii), we get
From equation (ii),
(b) When force is applied on heavier block B as shown in figure (c)
Divide (iv) by (iii), we get
From equation (iii)
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