(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.)
The answer to the first part is 16.4 W. I got 1.875, rounded it to 1.88 then to 1.87, and it still won't accept my answer. I only have 1 attempt to left. Please help. My online assignment is really conscientious about sig figs.
2.9*-3.2 +5.7*4.6 = 16.4 W yes
-1.7 * v = -6
v = 3.53 m/s
To solve part (a) of the problem, we need to calculate the instantaneous rate at which the force does work on the object.
The work done by a force can be calculated using the formula:
Work = Force dot Product Displacement
Since we are interested in the instantaneous rate at which the force does work on the object, we can calculate the dot product of the force (F) and the velocity (v) vectors. The dot product of two vectors is given by:
A dot B = (A_x * B_x) + (A_y * B_y) + (A_z * B_z)
Given:
F = 2.9 N i - 1.7 N j + 5.7 N k
v = -3.2 m/s i + 4.5 m/s k
Calculating the dot product of force and velocity:
F dot v = (2.9 N * -3.2 m/s) + (1.7 N * 0) + (5.7 N * 4.5 m/s)
= (-9.28 N*m/s) + (0 N*m/s) + (25.65 N*m/s)
= 16.37 N*m/s
Now, we have the work done by the force (F dot v) in units of N*m/s (or Joules per second). This is equivalent to the instantaneous power.
Therefore, the instantaneous rate at which the force does work on the object is 16.37 Watts (W).
Moving on to part (b) of the problem, we are given that the instantaneous power is -6.0 W. To find the velocity of the object at that time:
But since the force remains unchanged, we can equate the power to the dot product of the force and velocity vectors as we did in part (a):
-6.0 W = (2.9 N * V_x) + (1.7 N * 0) + (5.7 N * V_z)
Here, V_x and V_z represent the x and z components of the velocity, respectively.
Since there is no force along the y-direction and V_y is not given, we can assume it to be zero.
-6.0 W = (2.9 N * V_x) + (5.7 N * V_z)
You can solve this equation for either V_x or V_z using the given information, but it seems you're looking for V_x. Rearranging the equation:
2.9 N * V_x = -6.0 W - (5.7 N * V_z)
V_x = (-6.0 W - (5.7 N * V_z)) / 2.9 N
Plug in the given values for V_z and solve for V_x.
Remember to round your final answer to the appropriate number of significant figures based on the given values and the precision required by the problem.
Note: If you're still having trouble getting the correct answer, double-check all the calculations and make sure you're using the correct signs for the power and other values in the equations.
To find the instantaneous rate at which the force does work on the object, you can use the formula for work:
W = F · v
where W is the work done, F is the force, and v is the velocity.
(a) Given:
F = (2.9 N)i - (1.7 N)j + (5.7 N)k
v = -(3.2 m/s)i + (4.5 m/s)k
First, calculate the dot product of the force and velocity vectors:
F · v = (2.9 N)(-3.2 m/s) + (1.7 N)(0 m/s) + (5.7 N)(4.5 m/s)
Compute each term separately:
(2.9 N)(-3.2 m/s) = -9.28 N·m/s
(1.7 N)(0 m/s) = 0 N·m/s
(5.7 N)(4.5 m/s) = 25.65 N·m/s
Now sum up the terms:
F · v = -9.28 N·m/s + 0 N·m/s + 25.65 N·m/s = 16.37 N·m/s
Round the result to the appropriate number of significant figures, which is 3 in this case:
W = 16.4 N·m/s
So, the instantaneous rate at which the force does work on the object is 16.4 N·m/s.
(b) Given:
Power = -6.0 W
Power is related to work done per unit time:
Power = W / t
where W is the work done and t is the time interval.
However, since we are only given the instantaneous power and not the time interval, we need to find a different approach. We can use the relationship between power, force, and velocity:
Power = F · v
Substituting the given power value:
-6.0 W = (2.9 N)i - (1.7 N)j + (5.7 N)k · v
Since the given velocity only consists of a y-component, we can write it as:
v = 0i + vyj + 0k
Now, calculate the dot product:
-6.0 W = (2.9 N)(0) - (1.7 N)(vy) + (5.7 N)(0)
Simplifying:
0 = -1.7 N vy
Rearranging the equation:
vy = 0 m/s
So, the velocity of the object at that other time is 0 m/s in the y-direction.