A force acts on a 6.80 kg mobile object that moves from an initial position of to a final position of in 9.10 s. Find (a) the work done on the object by the force in the 9.10 s interval, (b) the average power due to the force during that interval, and (c) the angle between vectors and

a) 41.7 J

b) 4.6 W

c)

Explanation:

a)

The work done by a force on an object is given by

where

F is the force

d is the displacement of the object

is the scalar product

In this problem, the force is:

The initial position is

While the final position is

So, the displacement is the difference between r1 and r2:

And so, the work done is:

b)

The average power is calculated as the work done divided by the time taken:

where

W is the work done

t is the time taken to do that work

In this problem:

W = 41.7 J is the work done by the force

t = 9.10 s is the time taken for this work to be done

Therefore, the power used is:

c)

Given two vectors the angle between the two vectors can be found using

where

is the scalar product between u and v

is the magnitude of u

is the magnitude of v

Here the two vectors that we have are:

Their magnitudes are:

The scalar product is:

Therefore, the angle is:

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