Question

A force F = (3.15 N) + (7.00 N) + (7.05 N) acts on a 2.00 kg mobile object that moves from an initial position of di = (3.15 m) - (2.00 m) + (5.00 m) to a final position of df = -(5.00 m) + (3.90 m) + (7.00 m) in 4.00 s.

(a) Find the work done on the object by the force in the 4.00 interval J
(b) Find the average power due to the force during that interval. W
(c) Find the angle between vectors. di and df.°

Answer 1

To answer the student's question, work done is determined by the dot product of the force and displacement vectors, average power is work done divided by time, and the angle between di and df is calculated using the cosine of the dot product of vectors over the product of their magnitudes.

Step-by-step explanation:

The question involves calculating the work done, average power, and the angle between two vectors in the context of physics. Work is calculated as the dot product of the force vector and the displacement vector.

To find the angle between initial position (di) and final position (df) vectors, we use the dot product formula and the magnitude of vectors:

cos(θ) = (di · df) / (|di|*|df|)

Then, θ = cos^-1[(di · df) / (|di|*|df|)]

The amount of work done W is given by the formula:

W = F · Δd

Where F is the force vector and Δd the displacement vector (df - di).

The average power P during the time interval Δt can be found using the formula:

P = W / Δt