How Do You Calculate Work and Power for a Moving Object?

[danest]

Homework Statement

A force = (5.10 N) + (9.50 N) + (7.50 N) acts on a 2.30 kg mobile object that moves from an initial position of i = (1.30 m) - (9.70 m) + (7.90 m) to a final position of f = -(1.90 m) + (4.80 m) + (8.00 m) in 4.50 s. Find (a) the work done on the object by the force in the 4.50 s interval, (b) the average power due to the force during that interval, and (c) the angle between vectors i and f.

Homework Equations

W= F dot d

W= F*d

The Attempt at a Solution

First I attempted to solve for a by doing the dot product of the force vector and the d final but that didn't work. Then I found the magnitudes of F and d final and multiplied them but then that didn't work either. Am I supposed to use F= ma and find the acceleration? but if its that what am I supposed to find after?

Homework Statement

Homework Equations

The Attempt at a Solution

Solved

 

[rl.bhat]

Homework Statement

A force = (5.10 N) + (9.50 N) + (7.50 N) acts on a 2.30 kg mobile object that moves from an initial position of i = (1.30 m) - (9.70 m) + (7.90 m) to a final position of f = -(1.90 m) + (4.80 m) + (8.00 m) in 4.50 s. Find (a) the work done on the object by the force in the 4.50 s interval, (b) the average power due to the force during that interval, and (c) the angle between vectors i and f.

You better write the problem in the proper vector form. Like
F = (5.10 N)i + (9.50 N)j + (7.50 N)k
Initial position vector A = (1.30 m)i - (9.70 m)j + (7.90 m)k
Final position vector B = -(1.90 m)i + (4.80 m)j + (8.00 m)k
The displacement vector d = (B - A)
Now find w = F.d

 

[tomcue]

:

To solve this problem, we can use the equation W = F * d, where W is work, F is force, and d is displacement. We can also use the equation P = W/t, where P is power, t is time, and W is work.

(a) To find the work done on the object, we need to find the force and displacement vectors and then take their dot product. The force vector is given as (5.10 N) + (9.50 N) + (7.50 N) = (22.10 N) and the displacement vector is given as -(1.90 m) + (4.80 m) + (8.00 m) = (10.90 m). Taking the dot product, we get W = (22.10 N) * (10.90 m) = 240.29 J.

(b) To find the average power due to the force during the 4.50 s interval, we can use the equation P = W/t. We already know the value of W from part (a), and the time is given as 4.50 s. Plugging in the values, we get P = (240.29 J) / (4.50 s) = 53.40 W.

(c) To find the angle between the initial and final position vectors, we can use the dot product again. The dot product of two vectors A and B is given by A * B = |A| * |B| * cos(theta), where theta is the angle between the two vectors. In this case, A is the initial position vector and B is the final position vector. The magnitude of A is given by sqrt((1.30 m)^2 + (-9.70 m)^2 + (7.90 m)^2) = 13.47 m. The magnitude of B is given by sqrt((-1.90 m)^2 + (4.80 m)^2 + (8.00 m)^2) = 9.42 m. Plugging in the values, we get 13.47 m * 9.42 m * cos(theta) = (1.30 m) * (-1.90 m) + (-9.70 m) * (4.80 m) + (7.90 m) * (8.00 m). Solving for cos(theta), we get cos(theta) = 0.688