First slide

Application of Newtons laws

Question

The ratio of tensions in the string connected to the block of mass m₂ in Figure (a) and Figure (b) respectively, is (friction is absent everywhere): $[m_{1} = 50 kg, m_{2} = 80 kg$ $and F = 1000 N]$

Moderate

A

7:2
B

2:7
C

3:4
D

4:3

Solution

In Figure (a): Let tension in the string is T₁
$\begin{array}{l} F - T_{1} = m_{1} a \\ \Rightarrow 1000 - T_{1} = 50 a . . . . . . (i) \\ T_{1} - m_{2} g = m_{2} a \\ \Rightarrow T_{1} - 80 g = 80 a . . . . . . . (i i) \end{array}$
$From (i) and (ii) T_{1} = \frac{12000}{13} N$
In Figure (b): Let tension in the string is T₂.
$\begin{array}{l} F + m_{1} g - T_{2} = m_{1} a \\ \Rightarrow 1000 + 50 g - T_{2} = 50 a . . . . . . . (i i i) \\ T_{2} - 80 g = 80 a \\ T_{2} - 80 g = 80 a . . . . . . . (i v) \end{array}$
From (iii) and (iv),
$T_{2} = \frac{16000}{13} N \Rightarrow \frac{T_{1}}{T_{2}} = \frac{12000}{16000} = \frac{3}{4}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App