I'm having trouble with the following question. I got the answer 16.45 m/s² [up], which doesn't make sense to me, because that would mean the object is decelerating at a rate where the object would be travelling upwards. Please help me calculate the right answer (and show the steps) or explain to me why my answer is right if it is. Thanks.
A 30.0 gram object is thrown down with a velocity of 1.5 m/s. The air resistance constant is 0.35 (F = 0.35V²). Find the acceleration.
As a side question, is there any air resistance when the velocity is equal to 0? If I draw the free body diagram, would I only have to show that gravity is the only acting force?
A 30.0 gram object is thrown down with a velocity of 1.5 m/s. The air resistance constant is 0.35 (F = 0.35V²). Find the acceleration.
As a side question, is there any air resistance when the velocity is equal to 0? If I draw the free body diagram, would I only have to show that gravity is the only acting force?
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Use Newton's second law to find the instantaneous acceleration and choose up as the positive direction.
F net = ma
F (drag) - F (gravity) = ma
b v² - mg = ma
a = (b / m) v² - g
Now sub in the given numbers:
a = [(0.35 kg / m) / (0.030 kg)](1.5 m/s)² - 9.81 m/s²
a ≈ 16.44 m/s²
So your calculation was correct.
Since we have F (drag) = b v², when v = 0 we have F (drag) = 0, so there is no air resistance at this point (just like there is no kinetic friction on a stationary object.) I would only draw the gravitational force if you specify that it is for the special case in which v = 0.
Regarding your first question, the fact that the acceleration is large and positive does NOT mean that the object is travelling in the positive direction. The object was initially given a negative velocity, so it will be travelling downwards, but slowing down rapidly. However, as the velocity decreases, so too does the drag force, which means that it will never cause the object to actually travel upwards since the gravitational force is always present. After a long time has passed, the gravitational force will exactly equal the drag force (which can be shown by writing a = dv/dt, solving the resulting differential equation for v(t) and then taking the limit as t goes to infinity), and the "terminal velocity" will be attained:
F (grav) = F (drag)
mg = bv²
v = ± √(mg / b)
v ≈ ± 0.84 m/s
Is it plus or minus? Well of course it must be minus since the object was thrown downwards and gravity is acting downwards. Note that this analysis shows that the velocity of the object will never actually be 0.
Done!
F net = ma
F (drag) - F (gravity) = ma
b v² - mg = ma
a = (b / m) v² - g
Now sub in the given numbers:
a = [(0.35 kg / m) / (0.030 kg)](1.5 m/s)² - 9.81 m/s²
a ≈ 16.44 m/s²
So your calculation was correct.
Since we have F (drag) = b v², when v = 0 we have F (drag) = 0, so there is no air resistance at this point (just like there is no kinetic friction on a stationary object.) I would only draw the gravitational force if you specify that it is for the special case in which v = 0.
Regarding your first question, the fact that the acceleration is large and positive does NOT mean that the object is travelling in the positive direction. The object was initially given a negative velocity, so it will be travelling downwards, but slowing down rapidly. However, as the velocity decreases, so too does the drag force, which means that it will never cause the object to actually travel upwards since the gravitational force is always present. After a long time has passed, the gravitational force will exactly equal the drag force (which can be shown by writing a = dv/dt, solving the resulting differential equation for v(t) and then taking the limit as t goes to infinity), and the "terminal velocity" will be attained:
F (grav) = F (drag)
mg = bv²
v = ± √(mg / b)
v ≈ ± 0.84 m/s
Is it plus or minus? Well of course it must be minus since the object was thrown downwards and gravity is acting downwards. Note that this analysis shows that the velocity of the object will never actually be 0.
Done!