An object of mass 100g is thrown vertically upward from a point 60cm above the earths surface with an initial velocity of 150cm/sec. It rises briefly and then falls vertically to the earth, all of which time it is acted on by air resistance that is numerically equal to 200v (in dynes), where v is the velocity (in cm/sec).
a) find the velocity 0.1sec after the object is thrown.
b)find the velocity 0.1sec after the object stops rising and start falling.
net force=mass*acceleration
mg-200v=m*a
a=g-200v/mass
note we have a sign issue. if g is always downward, then if v is upowrd, then
a=g-200v/m
but when the motion is downward,
a=g+200v/m because v is now a negative sign....so on part b), one has to change the sign..
a) now we have an issue with calculus. One is tempted to use
vf=vi+at, however, a should be average acceleration, not the acceleration at t. However, because we are using very small time incrments, one can approximate the average a as the a at final time (similar to Simpson's right hand rule on incremental integration).
So as an approximation
vf=vi+at=150-(98+200v/100 )* .1
v=150-(98+.2v)
v(1.2)=52 calculate v
b. Same technique, however be mindful fo the sign for friction.
v(downward)=0-980-200v/100).1
v==98+.2v
v=-98/1.2 cm/sec
got this when i solve a)
ma = F
m dv/dt = -mg -kv
Putting in known quantities,
100 dv/dt = -100(980) - 200v
dv/dt = -980 - 2v
dv/dt + 2v = -980
v(0) = 150
IDK, TTYL, AF,BFF,SYl
To find the velocity of the object at different times, we can use the equations of motion along with the given initial conditions and the force of air resistance acting on the object.
a) To find the velocity 0.1 sec after the object is thrown, we need to consider the initial velocity, the gravitational force, and the air resistance force acting on the object.
The force of gravity acting on the object is given by:
F_gravity = m * g,
where m is the mass of the object (0.1kg) and g is the acceleration due to gravity (9.8m/s^2).
The air resistance force is given by:
F_air_resistance = 200v,
where v is the velocity of the object.
At any given time, the net force acting on the object is the difference between the force of gravity and the air resistance force:
F_net = F_gravity - F_air_resistance.
Using Newton's second law (F = ma), we can equate the net force to the product of the mass and acceleration:
F_net = m * a.
Combining the equations above, we have:
m * a = m * g - 200v.
Simplifying,
a = g - 200v.
Since the object is thrown vertically upward, the acceleration is in the opposite direction to gravity and will be negative (-9.8m/s^2).
Now, to find the velocity at 0.1s after the object is thrown, we can integrate the acceleration over time.
Integrating the equation a = g - 200v with respect to time (t) from initial velocity (u) to final velocity (v) and from 0 to 0.1 seconds, we get:
∫[0 to 0.1s] (dv) = ∫[0 to 0.1s] (g - 200v) dt,
v - u = [gt - 200(vt)] evaluated from 0 to 0.1s,
v - 150 = (9.8 * 0.1 - 200 * v * 0.1) - (0 - 0),
v - 150 = 0.98 - 20v.
Bringing terms involving 'v' to one side,
v + 20v = 0.98 + 150,
21v = 150.98,
v = 150.98 / 21 = 7.189 cm/s.
Therefore, the velocity 0.1 seconds after the object is thrown is approximately 7.189 cm/s.
b) To find the velocity 0.1 second after the object stops rising and starts falling, we need to consider the net force acting on the object. At the highest point of its trajectory, the velocity momentarily becomes zero.
At that point, the force of air resistance and the force due to gravity are both acting in the downward direction.
Using the same expressions for the net force and acceleration as in part (a), we have:
m * a = m * g - 200v.
Since the object has stopped rising and starts falling, the acceleration is in the downward direction and will be positive (9.8m/s^2).
Now, to find the velocity at 0.1s after the object starts falling, again, we can integrate the acceleration over time.
Integrating the equation a = g - 200v with respect to time (t) from initial velocity (u) to final velocity (v) and from 0 to 0.1 seconds, we get:
∫[0 to 0.1s] (dv) = ∫[0 to 0.1s] (g - 200v) dt,
v - u = ∫[0 to 0.1s] (9.8 - 200v) dt,
v - 0 = (9.8 * 0.1 - 200 * v * 0.1) - (0 - 0),
v = 0.98 - 20v.
Bringing terms involving 'v' to one side,
v + 20v = 0.98,
21v = 0.98,
v = 0.98 / 21 = 0.0467 m/s.
Therefore, the velocity 0.1 seconds after the object stops rising and starts falling is approximately 0.0467 m/s (or 4.67 cm/s).