Question

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

Answer 1

a) 41.7 J

b) 4.6 W

c)
`79.8^(\circ)`

Step-by-step explanation:

a)

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

`W=F\cdot d`

where

F is the force

d is the displacement of the object

`\cdot` is the scalar product

In this problem, the force is:

`F = (2.00i + 9.00j + 5.30k) N`

The initial position is

`r_1 = (2.70i - 2.90j + 5.50k)m`

While the final position is

`r_2 = (-4.10i + 3.30j + 5.40k) m`

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

`d=r_2-r_1=((-4.10-2.70)i+(3.30-(-2.90))j+(5.40-5.50)k)m =\\=(-6.80i+6.20j-0.10k)m`

And so, the work done is:

`W=F\cdot d=(2.00i+9.00j+5.30k)\cdot (-6.80i+6.20j-0.10k)=\\((2.00\cdot (-6.80))i+(9.00\cdot 6.20)j+(5.30\cdot (-0.10))k)=-13.6+55.8-0.5=\\41.7 J`

b)

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

`P=(W)/(t)`

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:

`P=(41.7)/(9.10)=4.6 W`

c)

Given two vectors
`u,v,` the angle between the two vectors can be found using

`cos \theta = (u\cdot v)/(|u||v|)`

where

`u\cdot v` is the scalar product between u and v

`|u|` is the magnitude of u

`|v|` is the magnitude of v

Here the two vectors that we have are:

`r_1 = (2.70i - 2.90j + 5.50k)m`

`r_2 = (-4.10i + 3.30j + 5.40k) m`

Their magnitudes are:

`|r_1|=√((2.70)^2+(-2.90)^2+(5.50)^2)=6.78 m`

`|r_2|=√((-4.10)^2+(3.30)^2+(5.40)^2)=7.54 m`

The scalar product is:

`r_1\cdot r_2=((2.70\cdot (-4.10))+((-2.90)\cdot 3.30)+(5.50\cdot 5.40))=9.06`

Therefore, the angle is:

`\theta=cos^(-1)((9.06)/((6.78)(7.54)))=79.8^(\circ)`