Respuesta :
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
[tex]s=r_{2}-r_{1}[/tex]
Where, [tex]r_{1}[/tex] = initial position
[tex]r_{2}[/tex] = final position
Put the value into the formula
[tex]s= (-4.10i + 3.30j + 5.40k)-(2.70i - 2.90j + 5.50k)[/tex]
[tex]s= -6.80i+6.20j-0.1k[/tex]
(a). We need to calculate the work done on the object
Using formula of work done
[tex]W=F\cdot s[/tex]
Put the value into the formula
[tex]W=(2.00i + 9.00j + 5.30k)\cdot (-6.80i+6.20j-0.1k)[/tex]
[tex]W=-13.6+55.8-0.53[/tex]
[tex]W=41.67\ J[/tex]
(b). We need to calculate the average power due to the force during that interval
Using formula of power
[tex]P=\dfrac{W}{t}[/tex]
Where, P = power
W = work
t = time
Put the value into the formula
[tex]P=\dfrac{41.67}{5.40}[/tex]
[tex]P=7.71\ Watt[/tex]
(c). We need to calculate the angle between vectors
Using formula of angle
[tex]\theta=\cos^{-1}(\dfrac{r_{1}r_{2}}{|r_{1}||r_{2}|})[/tex]
Put the value into the formula
[tex]\theta=\cos^{-1}\dfrac{(-4.10i + 3.30j + 5.40k)\cdot(2.70i - 2.90j + 5.50k)}{7.54\times6.778})[/tex]
[tex]\theta=79.7^{\circ}[/tex]
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°
