Question

A block of mass 1 kg is suspended by a string of mass 1 kg, length 1m as shown in figure(g=10m/s^2) Calculate:
(i) the tension in string at its lowest point.
(ii) the tension in string at its mid-point.
(iii) force Exerted by support on string.​

Answer 1

Answer:

Explanation:

Given :-

Mass of Block = 1 kg

Mass of String = 1 kg

Length of the string = 1 meter

Taking g = 10 m/s²

To Find :-

(a) The tension in string at its lowest point.

(b) The tension in string at its mid-point.

(c)  Force exerted by support on string.

Formula to be used :-

T = mg

Solution :-

(a) The tension in string at its lowest point.

Putting all the values, we get

⇒ T = mg

⇒ 1 × 10

⇒ T = 10 N.

(b) The tension in string at its mid-point.

At middle the mass of string gets half

Total mass = (1 + 0.5) kg = 1.5 kg

Tension at middle =  1.5 x 10 = 15 N.

(c)  Force exerted by support on string.

Total mass = 1 + 1 = 2 kg

Tension = 2 × 10  = 20 N

Answer 2

Answer:-

$T = 10 N \\ T_1 = 15 N \\T_2 = 20 N$

Given :-

$m = 1 kg \\l = 1 m \\ g = 10 m/s^2$

To find :-

(i) the tension in string at its lowest point.

(ii) the tension in string at its mid-point.

(iii) force Exerted by support on stri

Solution:-

• Take block as system.

Force acting on block :-

1) mg force downward

2) Tension force upward.

The system is in equilibrium.

→$\mathsf{ T = mg}$

→$\mathsf{ T = 1 \times 10}$

→$\mathsf{ T = 10 N}$

hence,

Tension force in string at lowest point is 10 N.

Case :- 2

Tension force at mid - point of the string.

L = m

L/2 = m /2

m = mass of block + mass of half string.

→$\mathsf{M = 1kg + 0.5 kg}$

→$\mathsf{ M = 1.5 kg}$

→$\mathsf{ T_1 = Mg}$

→$\mathsf{T_1 = 1.5 \times 10}$

→$\mathsf{ T_1 = 15 N}$

hence,

Tension force at middle of string is 15 N.

Case :- 3

Force exerted by support on string :-

Tension force by support = Tension force by string + mg force by block.

M" = mass of the block + mass of the string.

M" = 1 + 1

M" = 2 kg

→$\mathsf{T_2 = M"g}$

→$\mathsf{T_2 = 2 \times 10}$

→$\mathsf{ T_2 = 20 N}$

hence, tension force exerted by support on string is 20 N.