Answer:
-80 J
Explanation:
The first law of thermodynamics states that:
where
is the change in internal energy of the system
Q is the heat absorbed by the system
W is the work done by the system
In this problem, we have:
Q = -30 J is the heat released by the system (negative because the system releases it)
W = +50 J is the work done by the system on the surrounding (positive since it is done by the system)
Therefore, the change in internal energy is
The <em>estimated</em> displacement of the center of mass of the olive is ![\overrightarrow{\Delta r} = -0.046\,\hat{i} -0.267\,\hat{j}\,[m]](https://tex.z-dn.net/?f=%5Coverrightarrow%7B%5CDelta%20r%7D%20%3D%20-0.046%5C%2C%5Chat%7Bi%7D%20-0.267%5C%2C%5Chat%7Bj%7D%5C%2C%5Bm%5D)
.
<h3>Procedure - Estimation of the displacement of the center of mass of the olive</h3>
In this question we should apply the definition of center of mass and difference between the coordinates for <em>dynamic</em> (
) and <em>static</em> conditions (
) to estimate the displacement of the center of mass of the olive (
):
![\vec r - \vec r_{o} = \left[\frac{\Sigma\limits_{i=1}^{2}r_{i,x}\cdot(m_{i}\cdot g + F_{i, x})}{\Sigma \limits_{i =1}^{2}(F_{i,x}+m_{i}\cdot g)} ,\frac{\Sigma\limits_{i=1}^{2}r_{i,y}\cdot(m_{i}\cdot g + F_{i, y})}{\Sigma \limits_{i =1}^{2}(F_{i,y}+m_{i}\cdot g)} \right]-\left(\frac{\Sigma\limits_{i=1}^{2}r_{i,x}\cdot m_{i}\cdot g}{\Sigma \limits_{i= 1}^{2} m_{i}\cdot g}, \frac{\Sigma\limits_{i=1}^{2}r_{i,y}\cdot m_{i}\cdot g}{\Sigma \limits_{i= 1}^{2} m_{i}\cdot g}\right)](https://tex.z-dn.net/?f=%5Cvec%20r%20-%20%5Cvec%20r_%7Bo%7D%20%3D%20%5Cleft%5B%5Cfrac%7B%5CSigma%5Climits_%7Bi%3D1%7D%5E%7B2%7Dr_%7Bi%2Cx%7D%5Ccdot%28m_%7Bi%7D%5Ccdot%20g%20%2B%20F_%7Bi%2C%20x%7D%29%7D%7B%5CSigma%20%5Climits_%7Bi%20%3D1%7D%5E%7B2%7D%28F_%7Bi%2Cx%7D%2Bm_%7Bi%7D%5Ccdot%20g%29%7D%20%2C%5Cfrac%7B%5CSigma%5Climits_%7Bi%3D1%7D%5E%7B2%7Dr_%7Bi%2Cy%7D%5Ccdot%28m_%7Bi%7D%5Ccdot%20g%20%2B%20F_%7Bi%2C%20y%7D%29%7D%7B%5CSigma%20%5Climits_%7Bi%20%3D1%7D%5E%7B2%7D%28F_%7Bi%2Cy%7D%2Bm_%7Bi%7D%5Ccdot%20g%29%7D%20%5Cright%5D-%5Cleft%28%5Cfrac%7B%5CSigma%5Climits_%7Bi%3D1%7D%5E%7B2%7Dr_%7Bi%2Cx%7D%5Ccdot%20m_%7Bi%7D%5Ccdot%20g%7D%7B%5CSigma%20%5Climits_%7Bi%3D%201%7D%5E%7B2%7D%20m_%7Bi%7D%5Ccdot%20g%7D%2C%20%5Cfrac%7B%5CSigma%5Climits_%7Bi%3D1%7D%5E%7B2%7Dr_%7Bi%2Cy%7D%5Ccdot%20m_%7Bi%7D%5Ccdot%20g%7D%7B%5CSigma%20%5Climits_%7Bi%3D%201%7D%5E%7B2%7D%20m_%7Bi%7D%5Ccdot%20g%7D%5Cright%29)
(1)
Where:
- x-Coordinate of the i-th element of the system, in meters.
- y-Coordinate of the i-th element of the system, in meters. 
- x-Component of the net force applied on the i-th element, in newtons.
- y-Component of the net force applied on the i-th element, in newtons. 
- Mass of the i-th element, in kilograms.
- Gravitational acceleration, in meters per square second.
If we know that ![\vec r_{1} = (0, 0)\,[m]](https://tex.z-dn.net/?f=%5Cvec%20r_%7B1%7D%20%3D%20%280%2C%200%29%5C%2C%5Bm%5D)
, ![\vec r_{2} = (1, 2)\,[m]](https://tex.z-dn.net/?f=%5Cvec%20r_%7B2%7D%20%3D%20%281%2C%202%29%5C%2C%5Bm%5D)
, ![\vec F_{1} = (0, 3)\,[N]](https://tex.z-dn.net/?f=%5Cvec%20F_%7B1%7D%20%3D%20%280%2C%203%29%5C%2C%5BN%5D)
, ![\vec F_{2} = (-3, -2)\,[N]](https://tex.z-dn.net/?f=%5Cvec%20F_%7B2%7D%20%3D%20%28-3%2C%20-2%29%5C%2C%5BN%5D)
, 
, 
and 
, then the displacement of the center of mass of the olive is:
<h3>Dynamic condition
![\vec{r} = \left[\frac{(0)\cdot (0.50)\cdot (9.807)+(0)\cdot (0) + (1)\cdot (1.50)\cdot (9.807) + (1)\cdot (-3)}{(0.50)\cdot (9.807) + 0 + (1.50)\cdot (9.807)+(-3)}, \frac{(0)\cdot (0.50)\cdot (9.807) + (0)\cdot (3) + (2)\cdot (1.50)\cdot (9.807) +(2) \cdot (-2)}{(0.50)\cdot (9.807) + (3)+(1.50)\cdot (9.807)+(-2)} \right]](https://tex.z-dn.net/?f=%5Cvec%7Br%7D%20%3D%20%5Cleft%5B%5Cfrac%7B%280%29%5Ccdot%20%280.50%29%5Ccdot%20%289.807%29%2B%280%29%5Ccdot%20%280%29%20%2B%20%281%29%5Ccdot%20%281.50%29%5Ccdot%20%289.807%29%20%2B%20%281%29%5Ccdot%20%28-3%29%7D%7B%280.50%29%5Ccdot%20%289.807%29%20%2B%200%20%2B%20%281.50%29%5Ccdot%20%289.807%29%2B%28-3%29%7D%2C%20%5Cfrac%7B%280%29%5Ccdot%20%280.50%29%5Ccdot%20%289.807%29%20%2B%20%280%29%5Ccdot%20%283%29%20%2B%20%282%29%5Ccdot%20%281.50%29%5Ccdot%20%289.807%29%20%2B%282%29%20%5Ccdot%20%28-2%29%7D%7B%280.50%29%5Ccdot%20%289.807%29%20%2B%20%283%29%2B%281.50%29%5Ccdot%20%289.807%29%2B%28-2%29%7D%20%20%5Cright%5D)
![\vec r = (0,704, 1.233)\,[m]](https://tex.z-dn.net/?f=%5Cvec%20r%20%3D%20%280%2C704%2C%201.233%29%5C%2C%5Bm%5D)
</h3>
<h3>Static condition</h3><h3>
![\vec{r}_{o} = \left[\frac{(0)\cdot (0.50)\cdot (9.807) + (1)\cdot (1.50)\cdot (9.807)}{(0.50)\cdot (9.807) + (1.50)\cdot (9.807)}, \frac{(0)\cdot (0.50)\cdot (9.807) + (2)\cdot (1.50)\cdot (9.807)}{(0.50)\cdot (9.807)+(1.50)\cdot (9.807)} \right]](https://tex.z-dn.net/?f=%5Cvec%7Br%7D_%7Bo%7D%20%3D%20%5Cleft%5B%5Cfrac%7B%280%29%5Ccdot%20%280.50%29%5Ccdot%20%289.807%29%20%2B%20%281%29%5Ccdot%20%281.50%29%5Ccdot%20%289.807%29%7D%7B%280.50%29%5Ccdot%20%289.807%29%20%2B%20%281.50%29%5Ccdot%20%289.807%29%7D%2C%20%5Cfrac%7B%280%29%5Ccdot%20%280.50%29%5Ccdot%20%289.807%29%20%2B%20%282%29%5Ccdot%20%281.50%29%5Ccdot%20%289.807%29%7D%7B%280.50%29%5Ccdot%20%289.807%29%2B%281.50%29%5Ccdot%20%289.807%29%7D%20%20%5Cright%5D)
</h3><h3>
![\vec r_{o} = \left(0.75, 1.50)\,[m]](https://tex.z-dn.net/?f=%5Cvec%20r_%7Bo%7D%20%3D%20%5Cleft%280.75%2C%201.50%29%5C%2C%5Bm%5D)
</h3><h3 /><h3>Displacement of the center of mass of the olive</h3>
![\overrightarrow{\Delta r} = (0.704-0.75, 1.233-1.50)\,[m]](https://tex.z-dn.net/?f=%5Coverrightarrow%7B%5CDelta%20r%7D%20%3D%20%280.704-0.75%2C%201.233-1.50%29%5C%2C%5Bm%5D)
![\overrightarrow{\Delta r} = (-0.046, -0.267)\,[m]](https://tex.z-dn.net/?f=%5Coverrightarrow%7B%5CDelta%20r%7D%20%3D%20%28-0.046%2C%20-0.267%29%5C%2C%5Bm%5D)
The <em>estimated</em> displacement of the center of mass of the olive is . 
To learn more on center of mass, we kindly invite to check this verified question: brainly.com/question/8662931
If the wagon travels 18.75 m, then the work done on the wagon is
(18.75 m) x (the steady force applied to the wagon all the way, in Newtons) .
The unit is Joules .
