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

satendra , 11 Years ago
Grade 12th pass
anser 2 Answers
bharat bajaj

Last Activity: 11 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

Last Activity: 8 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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