a 5.00 kg block is pulled along a horizontal frictionless floor by a cord that exerts a force of magnitude F=12.0 N at an anglr of 25 degrees above the horizontal. What is the magnitude of the block's acceleration? the force magnitude F slowly increased. What is its value just before the block is lifted (completely) off the floor?
what is the magnitude of the block's acceleration just before is lifted off the floor (completely.
2007-11-07 03:38:26 · 4 answers · asked by magna1211@sbcglobal.net 1 in Science & Mathematics ➔ Physics
The horizontal component of the force is given by:
Fx = F cos theta
The vertical component is canceled by the normal force, so because there's no friction, it's unimportant to the motion (until liftoff).
Fx = ma, so a = Fx / m = F cos (theta) / m.
They give you F, m, and theta.
The block is lifted when the vertical component exceeds the weight, ie
F sin theta = mg
so F = mg / sin (theta)
You can calculate the acceleration with that force:
a = Fx / m = F cos theta / m = g cot (theta)
2007-11-07 03:50:03 · answer #1 · answered by Anonymous · 0⤊ 0⤋
In the first part of the question, F is given as 12 N.
Hence the acceleration
a = Force in the direction of accn. / Mass = 12 cos 25 / 5
= 0.3625 m/s^2.
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In the second part of the question the value of F is not given.
We have to find it by the given condition.
It is given the box is just lifted of the ground.
Therefore its weight must be balanced by the vertical component of the force F
F sin 25 = 5* 9.8.
From this we find F as equal to 115.94 N
Now we can do the problem as done in the first part
a = 115.94 cos 25 / 5 = 21 m/s^2.
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Having understood the concept we can simply find the acceleration by using
a = 9.8 / tan 25 = 21 m/ s^2.
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2007-11-07 12:12:00 · answer #2 · answered by Pearlsawme 7 · 0⤊ 0⤋
Horizontal force = CAH = cos 25 * 12 = 10.9
F = MA, so 10.9 = 5a; 10.9/5 = a;
Start acceleration = 2.175 m/s/s
Vertical force must be greater than or equal to 9.8*5 (gravity * mass) = 49 N
Vertical Force = SIN 25 * F = 49
49/SIN 25 = F
End Force = 116 N
Horizontal force = TOA = 116/TAN 25
248.6
F = MA
248.6 = 5a
End Acceleration = 49.7
2007-11-07 11:54:16 · answer #3 · answered by Anonymous · 0⤊ 0⤋
F(hor) = 12.0cos25 = 10.9 N
a(block) = F/m = 10.9/5.00 = 2.18 m/sec²
F(lift) = 5.00*9.8/sin25 = 116 N
a(block at lift) = 116cos25/5.00 = 21.0 m/sec²
(all answers to 3 sig figs)
2007-11-07 11:50:47 · answer #4 · answered by gebobs 6 · 0⤊ 0⤋