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# Mechanics (M1) question watch

1. can someone give me a hand with this question i am completely lost!

A box of mass 5kg is being pulled along a rough horizontal ground by a rope inclined at 20 degrees to the horizontal. When the tension in the rope is 10 N the box is moving at a constant speed.

A) calculate the coefficient of friction between the box and the ground.

b) The tension in the rope is now 15 N
calculate the acceleration of the box
2. 10N = limiting friction and weight acting down the slope
10 = F + 5gsin20
F = 10 - 5gsin20
normal reaction = 5gcos20
(mu) = (10 - 5gsin20)/5gcos20

F = ma
15 - (10 - 5gsin20) - 5gsin20 = 5a
15-10 = 5a
a = 1 ms-2
3. (Original post by Jutx)
can someone give me a hand with this question i am completely lost!

A box of mass 5kg is being pulled along a rough horizontal ground by a rope inclined at 20 degrees to the horizontal. When the tension in the rope is 10 N the box is moving at a constant speed.

A) calculate the coefficient of friction between the box and the ground.

b) The tension in the rope is now 15 N
calculate the acceleration of the box
Resolve horizontally:
10cos20-UR=0
10cos20-5gU=
5gU=10cos20
U=10cos20/5g
=10cos20/49
=0.192

b) 15cos20- (0.192x49)=5a
4.687=5a
a=0.94ms-2
4. cheers. I am not the best person at mechanics lol
5. (Original post by Undry1)
10N = limiting friction and weight acting down the slope
10 = F + 5gsin20
F = 10 - 5gsin20
normal reaction = 5gcos20
(mu) = (10 - 5gsin20)/5gcos20

F = ma
15 - (10 - 5gsin20) - 5gsin20 = 5a
15-10 = 5a
a = 1 ms-2
Horziontal rough plane
6. (Original post by Malik)
Horziontal rough plane
oops.
7. (Original post by Jutx)
can someone give me a hand with this question i am completely lost!

A box of mass 5kg is being pulled along a rough horizontal ground by a rope inclined at 20 degrees to the horizontal. When the tension in the rope is 10 N the box is moving at a constant speed.

A) calculate the coefficient of friction between the box and the ground.

b) The tension in the rope is now 15 N
calculate the acceleration of the box
R = upwards force = 10sin20 + s

vertically: R = 10sin20 + s = 5g
horizontally: F = µR = 10cos20

µ = 10cos20/5g = 0.19...

b) F = ma
a = F/m = (15cos20 - µ5g)/5 = 0.93ms^-2 (which is cos20, don't you know)

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Updated: November 13, 2005
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