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Grade: 12th pass


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 is a) 1.5 m/sec. b) 60 m/sec. c)0.1 m/sec. d) 5 m/sec.

6 years ago

Answers : (2)

bharat bajaj
IIT Delhi
askIITians Faculty
122 Points
							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

Bharat Bajaj
Askiitians faculty
6 years ago
11 Points
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
4 years ago
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