Question

A force F=(3.00N)i+(7.00N)j+(7.00N)k acts on a 2.00kg mobile object that moves from an initial position of di​=(−3.00m)i−(2.00m)j−(5.00m)k to a final position df​=(5.00m)i+(4.00m)j+(7.00m)k in 4.00s. Find:

The work done on the object by the force in the 4.00s interval.

Answer 1

The work done by the force on the object is calculated using the dot product of the force and displacement vectors, providing a solution of 150.00 Joules.

Step-by-step explanation:

Work Done by a Force

To find the work done by the force F on the object during the 4.00s interval, we need to calculate the dot product of the force vector F and the displacement vector Δd. The force vector is given as F = (3.00N)i + (7.00N)j + (7.00N)k, and the displacement Δd is found by subtracting the initial position vector di from the final position vector df, which yields Δd = df - di = (5.00m)i + (4.00m)j + (7.00m)k - ((-3.00m)i - (2.00m)j - (5.00m)k) = (8.00m)i + (6.00m)j + (12.00m)k.

The work done W can be calculated as:
W = F · Δd = (3.00N)(8.00m) + (7.00N)(6.00m) + (7.00N)(12.00m)

By solving the mathematical problem completely, we get:
W = 24.00 + 42.00 + 84.00 = 150.00 J

Therefore, the work done by the force on the object over the 4.00s interval is 150.00 Joules.