A rope of length 10 m and linear density 0.5 kg m- 1 is lying length wise on a smooth horizontal floor. It is pulled by a force of 25 N . The tension in the rope at a point 6 m away from the point of application is

Question

A rope of length 10 m and linear density 0.5 kg m- 1 is lying length wise on a smooth horizontal floor. It is pulled by a force of 25 N . The tension in the rope at a point 6 m away from the point of application is (A) 20 N (B) 15 N (C) 10 N (D) 5 N

Tardigrade Admin · Accepted Answer

Mass of rope =10× 0.5=5 kg Given force =25 N Acceleration =(F/m)=(25/5)=5 ms- 2 Length of remaining rope =4m Hence mass of remaining rope =4× 0.5=2kg Hence tension on the rope at a point 6m away =m× a=2× 5=10N

Q. A rope of length $10 m$ and linear density $0.5 k g m^{- 1}$ is lying length wise on a smooth horizontal floor. It is pulled by a force of $25 N$ . The tension in the rope at a point $6 m$ away from the point of application is

20 N

15 N

10 N

5 N

Mass of rope $= 10 \times 0.5 = 5 k g$
Given force $= 25 N$
Acceleration $= \frac{F}{m} = \frac{25}{5} = 5 m s^{- 2}$
Length of remaining rope $= 4 m$
Hence mass of remaining rope $= 4 \times 0.5 = 2 k g$
Hence tension on the rope at a point 6m away $= m \times a = 2 \times 5 = 10 N$

Q. A rope of length 10m and linear density 0.5kgm−1 is lying length wise on a smooth horizontal floor. It is pulled by a force of 25N . The tension in the rope at a point 6m away from the point of application is

20 N

15 N

10 N

5 N

Mass of rope =10×0.5=5kg Given force =25N Acceleration =mF​=525​=5ms−2 Length of remaining rope =4m Hence mass of remaining rope =4×0.5=2kg Hence tension on the rope at a point 6m away =m×a=2×5=10N

Q. A rope of length $10 m$ and linear density $0.5 k g m^{- 1}$ is lying length wise on a smooth horizontal floor. It is pulled by a force of $25 N$ . The tension in the rope at a point $6 m$ away from the point of application is

Mass of rope $= 10 \times 0.5 = 5 k g$
Given force $= 25 N$
Acceleration $= \frac{F}{m} = \frac{25}{5} = 5 m s^{- 2}$
Length of remaining rope $= 4 m$
Hence mass of remaining rope $= 4 \times 0.5 = 2 k g$
Hence tension on the rope at a point 6m away $= m \times a = 2 \times 5 = 10 N$