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# A ship of mass 3 × 107 kg initially at rest, is pulled by a force of 5 × 104 N through a distance of 3m. Assuming that the resistance due to water is negligible, the speed of the ship isa) 1.5 m/sec.b) 60 m/sec.c)0.1 m/sec.d) 5 m/sec.

Grade:12th pass

## 2 Answers

bharat bajaj IIT Delhi
askIITians Faculty 122 Points
7 years ago
A = F/m = 5 × 104/ 3x107
From newton third law
V^2 - u^2 = 2as
V^2 = 2 x 5 x 3 x 104/3x107
It comes out to be near 3 m/ sec
There is no such option. The answer is correct. Please check the values again

Thanks
Bharat Bajaj
Askiitians faculty
Nilesh
11 Points
4 years ago
F=ma
5*104 = (3*107) a
a = (5/3)*10-3

Now, By newton’s third law of motion,
v2 – u2 =2as
v2 – 02 = (2)*(5/3)*(10-3)(3)
v2 = 10-2
v = 0.1 m/s

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