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Help with qn 7 pls!

Date Posted: 3 years ago
Views: 388
Eric Nicholas K
Eric Nicholas K
3 years ago
Q7 I am not too sure, but I believe that the force applied acts against two things. Work done against gravity (which is “mgh”) and work done against friction (on a rough surface).

Q8 the answer should be A.
Eric Nicholas K
Eric Nicholas K
3 years ago
So based on my line of thought,

Work done by the pulling = “F x parallel d” regardless of whether the block is moving horizontally or up a slope = 100 N x 8 m = 800 J

This is counteracted by the work gone against gravity = 20 kg x 10 N/kg x 3 m = 600 J...

...and work done against friction of 200 J.
Eric Nicholas K
Eric Nicholas K
3 years ago

Assuming no energy is lost, all the kinetic energy is converted to gravitational potential energy.

0.5 mv2 = mgh

Assuming the bullet’s mass remains unchanged throughout (well, sometimes it disintegrates slightly), the equation simplifies to

v2 = 2gh = 2 x 10 N/kg x 720 m = 14400 m2/s2

v = square root of that answer = 120 m/s

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