(a) At a certain instant, a particle-like object is acted on by a force F=(2.9N)i-(1.7N)j+(5.7N)k while the object's velocity is v= -(3.2m/s)i+(4.5m/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 y component. If the force is unchanged, and the instantaneous power is -6.0 W, what is the velocity of the object just then? (Give your answer without a unit vector.)
(a) To find the instantaneous rate at which the force does work on the object, we can use the equation:
Power = Force dot Product Velocity
The dot product of two vectors is given by:
A dot B = A * B cos(theta)
where A and B are vectors and theta is the angle between them.
In this case, the force vector F = (2.9 N)i - (1.7 N)j + (5.7 N)k and the velocity vector v = -(3.2 m/s)i + (4.5 m/s)k.
Calculating the dot product, we get:
F dot v = (2.9 N * -3.2 m/s) + (-1.7 N * 0 m/s) + (5.7 N * 4.5 m/s) = -9.28 N m/s + 25.65 N m/s = 16.37 N m/s
Therefore, the instantaneous rate at which the force does work on the object is 16.37 N m/s.
(b) Given that the force is unchanged and the instantaneous power is -6.0 W, we can use the equation:
Power = Force dot Product Velocity
We know the power is -6.0 W, and the force vector F remains the same. Let's assume the velocity vector at this time is v = (0 m/s)i + vy j + (0 m/s)k, where vy is the y-component of the velocity.
The dot product becomes:
F dot v = (2.9 N * 0 m/s) + (-1.7 N * vy m/s) + (5.7 N * 0 m/s) = -1.7 N vy m/s
Since we know that F dot v is equal to -6.0 W, we can set up the equation:
-1.7 N vy m/s = -6.0 W
To find vy, we divide both sides by -1.7 N:
vy = -6.0 W / -1.7 N ≈ 3.53 m/s
Therefore, the velocity of the object at that time is approximately (0 m/s)i + 3.53 m/sj + (0 m/s)k.
(a) The instantaneous rate at which the force does work on the object can be calculated using the formula:
Power = Force · Velocity
Given that the force F = (2.9N)i - (1.7N)j + (5.7N)k and the velocity v = -(3.2m/s)i + (4.5m/s)k, we can substitute these values into the formula:
Power = (2.9N)i · (-(3.2m/s)i) + (2.9N)i · (4.5m/s)k + (-1.7N)j · (-(3.2m/s)i) + (-1.7N)j · (4.5m/s)k + (5.7N)k · (-(3.2m/s)i) + (5.7N)k · (4.5m/s)k
Simplifying this equation, we get:
Power = -9.28 N·m/s + (-7.65 N·m/s) + (-18.24 N·m/s) + (25.65 N·m/s)
Now, we can sum up these values:
Power = -9.28 N·m/s - 7.65 N·m/s - 18.24 N·m/s + 25.65 N·m/s
Power = 0.48 N·m/s
Therefore, the instantaneous rate at which the force does work on the object is 0.48 N·m/s.
(b) In this case, since the force is unchanged, and the instantaneous power is given as -6.0 W, we can use the formula:
Power = Force · Velocity
Again, substituting the values:
-6.0 W = (2.9N)i · (v_y) + (-1.7N)j · (v_y) + (5.7N)k · (v_y)
Since the force has components only in the x and z directions, and the velocity at this time only has a y component, we can deduce that the force and velocity have no overlap in these components. Therefore, we can ignore the y component of the force and velocity, resulting in:
-6.0 W = (5.7N)k · (v_y)
Simplifying, we get:
-6.0 W = (5.7N) v_y
Dividing both sides by 5.7N:
v_y = -6.0 W / 5.7N
v_y = -1.05 m/s
Therefore, the velocity of the object at that time is -1.05 m/s in the y direction.