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) $79.8^{\circ}$

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:

$P=\frac{W}{t}$

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:

$P=\frac{41.7}{9.10}=4.6 W$

c)

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

$cos \theta = \frac{u\cdot v}{|u||v|}$

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:

$|r_1|=\sqrt{(2.70)^2+(-2.90)^2+(5.50)^2}=6.78 m$

$|r_2|=\sqrt{(-4.10)^2+(3.30)^2+(5.40)^2}=7.54 m$

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:

$\theta=cos^{-1}(\frac{9.06}{(6.78)(7.54)})=79.8^{\circ}$