At a certain instant, a particle-like object is acted on by a force F = (3.0 N) i - (1.0 N) j + (8.0 N) k while the object's velocity is v = - (2.0 m/s) i + (4.0 m/s) k. What is the instantaneous rate at which the force does work on the object?
(b) At some other time, the velocity consists of only a j component. If the force is unchanged, and the instantaneous power is -12 W, what is the velocity of the object just then?
a) the Dot product of force and velocity is power.
Well, well, well! Looks like we have a particle-like object getting all worked up by a force! I'm here to help you make sense of it all, with a side of silliness, of course!
(a) To find the instantaneous rate at which the force does work on the object, we need to use the dot product of the force and velocity. It's like a fancy mathematical game of tag! Ready? Let's go!
Force (F) = 3.0 N i - 1.0 N j + 8.0 N k
Velocity (v) = -2.0 m/s i + 4.0 m/s k
Now, we just need to find the dot product of these two vectors and *POOF*, we'll have the answer!
F dot v = (3.0 N * -2.0 m/s) + (-1.0 N * 0) + (8.0 N * 4.0 m/s)
But hey, remember that anything multiplied by 0 is 0, so we get:
F dot v = -6.0 N m/s + 0 + 32.0 N m/s
Add those up, and we have:
F dot v = 26.0 N m/s (TA-DAH!)
So, the instantaneous rate at which the force does work on the object is 26.0 N m/s. Nice work, team!
(b) Alrighty, let's move on to the next stop on our wacky physics ride! If the object's velocity consists of only a j component, and the force remains unchanged, we're on a mission to find the velocity at just that moment. Buckle up!
Instantaneous power (P) = -12 W
Pssst, I'll let you in on a little secret. Power (P) is also the dot product of force (F) and velocity (v) divided by time (t). But here's the kicker, we're given power and want to find velocity! Time to channel our mathematical mojo:
P = F dot v / t
Since we want to find velocity (v), we can rearrange the equation like this:
v = P * t / F
But wait, we were also given the force F = (3.0 N) i - (1.0 N) j + (8.0 N) k. So, let's substitute that in:
v = P * t / (3.0 N i - 1.0 N j + 8.0 N k)
And there you have it, my friend! Just plug in the values for power (P), time (t), and the components of force (F), and you'll uncover the velocity at that very moment. Clowns honor!
Remember, my explanations may be funny, but the math is serious business. So, have fun crunching those numbers and enjoy the journey!
To find the instantaneous rate at which the force does work on the object, we first need to find the dot product of the force and velocity.
(a) The dot product of two vectors A and B can be calculated using the formula A · B = |A| |B| cosθ, where θ is the angle between the two vectors.
Given:
Force F = (3.0 N) i - (1.0 N) j + (8.0 N) k
Velocity v = - (2.0 m/s) i + (4.0 m/s) k
The dot product of F and v is: F · v = (3.0 N * -2.0 m/s) + (-1.0 N * 4.0 m/s) + (8.0 N * 0)
= -6.0 Nm/s - 4.0 Nm/s + 0 Nm/s
= -10.0 Nm/s
Therefore, the instantaneous rate at which the force does work on the object is -10.0 Nm/s.
(b) To find the velocity of the object at a different time, given the force is unchanged and the instantaneous power is -12 W, we can use the formula for power:
Power = F · v
Given power = -12 W
Force F = (3.0 N) i - (1.0 N) j + (8.0 N) k
Let's find the dot product of F and v:
-12 W = (3.0 N * v_x) + (-1.0 N * v_y) + (8.0 N * v_z)
Since the velocity consists of only a j component, we can write v = 0i + v_yj + 0k.
Plugging this into the equation, we get:
-12 W = (3.0 N * 0) + (-1.0 N * v_y) + (8.0 N * 0)
-12 W = -1.0 N * v_y
Solving for v_y:
v_y = -12 W / -1.0 N
v_y = 12 m/s
Therefore, the velocity of the object at that time is v = 0i + 12j + 0k.
To find the instantaneous rate at which the force does work on the object, we need to calculate the dot product of the force and the velocity.
The dot product of two vectors is given by the formula: A · B = |A| |B| cosθ, where A and B are the vectors, |A| and |B| are their magnitudes, and θ is the angle between them.
In this case, the force (F) is given as (3.0 N)i - (1.0 N)j + (8.0 N)k, and the velocity (v) is given as - (2.0 m/s)i + (4.0 m/s)k.
To calculate the dot product, we multiply the respective components of the force and velocity vectors and then sum them up:
F · v = (3.0 N)(-2.0 m/s) + (-1.0 N)(0) + (8.0 N)(4.0 m/s)
Since the j components of the force and velocity vectors are perpendicular, the dot product of the j components is zero. Therefore, the calculation simplifies to:
F · v = (3.0 N)(-2.0 m/s) + (8.0 N)(4.0 m/s)
Now, we can perform the calculations:
F · v = -6.0 N m/s + 32.0 N m/s
F · v = 26.0 N m/s
Therefore, the instantaneous rate at which the force does work on the object is 26.0 N m/s.
Now, let's move on to the second part of the question.
In this part, we are given that the instantaneous power is -12W and the force remains unchanged. We need to find the velocity of the object at that time.
The instantaneous power is given by the formula: P = F · v, where P is power and F and v are vectors.
Rearranging the formula, we can solve for v:
v = P / F
Substituting the given values:
v = -12W / (3.0 N)i - (1.0 N)j + (8.0 N)k
We divide each component of the force vector by its respective magnitude to obtain the unit vectors:
v = -12W / (3.0 N)(-2.0 m/s)i + (1.0 N)(0)m/sj + (8.0 N)(4.0 m/s)k
Now, we can perform the calculations:
v = -12W / (-6.0 N m/s)i + 0 m/sj + 32.0 N m/sk
v = 2.0 m/s i + 0 m/s j - 4.0 m/s k
Therefore, the velocity of the object at that time is 2.0 m/s i - 4.0 m/s k.