A force acts on a 9.90 kg mobile object that moves from an initial position of to a final position of in 5.40 s. Find (a) the work done on the object by the force in the 5.40 s interval, (b) the average power due to the force during that interval, and (c) the angle between vectors and .

Given that,

Mass of object = 9.90 kg

Time =5.40 s

Suppose the force is (2.00i + 9.00j + 5.30k) N, initial position is (2.70i - 2.90j + 5.50k) m and final position is (-4.10i + 3.30j + 5.40k) m.

We need to calculate the displacement

Using formula of displacement

Where, = initial position

= final position

Put the value into the formula

(a). We need to calculate the work done on the object

Using formula of work done

Put the value into the formula

(b). We need to calculate the average power due to the force during that interval

Using formula of power

Where, P = power

W = work

t = time

Put the value into the formula

(c). We need to calculate the angle between vectors

Using formula of angle

Put the value into the formula

Hence, (a). The work done on the object by the force in the 5.40 s interval is 41.67 J.

(b). The average power due to the force during that interval is 7.71 Watt.

(c).  The angle between vectors is 79.7°

the answer is b i think