The components of a force acting on a particl | vedclass

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(C) From the given graphs,we determine the equations for the force components:$1$. For $F_x$ vs $x$: The slope is $	an(37^\circ) = 3/4$. The intercep

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$250$

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(C) From the given graphs,we determine the equations for the force components:$1$. For $F_x$ vs $x$: The slope is $	an(37^\circ) = 3/4$. The intercept is $10$. Thus,$F_x = \frac{3}{4}x + 10$.$2$. For $F_y$ vs $y$: The line passes through $(0, 20)$ and $(15, 0)$. The slope is $(0-20)/(15-0) = -4/3$. Thus,$F_y = -\frac{4}{3}y + 20$.$3$. For $F_z$ vs $z$: The line passes through $(0, -16)$ and $(12, 0)$. The slope is $(0 - (-16))/(12-0) = 16/12 = 4/3$. Thus,$F_z = \frac{4}{3}z - 16$.The work done is $W = \int \vec{F} \cdot d\vec{r} = \int_{x_A}^{x_B} F_x dx + \int_{y_A}^{y_B} F_y dy + \int_{z_A}^{z_B} F_z dz$.$W = \int_{0}^{8} (\frac{3}{4}x + 10) dx + \int_{5}^{20} (-\frac{4}{3}y + 20) dy + \int_{12}^{0} (\frac{4}{3}z - 16) dz$.Calculating each integral:$W_x = [\frac{3}{8}x^2 + 10x]_0^8 = \frac{3}{8}(64) + 10(8) = 24 + 80 = 104 \ J$.$W_y = [-\frac{2}{3}y^2 + 20y]_5^{20} = (-\frac{2}{3}(400) + 400) - (-\frac{2}{3}(25) + 100) = (-\frac{800}{3} + \frac{1200}{3}) - (-\frac{50}{3} + \frac{300}{3}) = \frac{400}{3} - \frac{250}{3} = \frac{150}{3} = 50 \ J$.$W_z = [\frac{2}{3}z^2 - 16z]_{12}^0 = (0) - (\frac{2}{3}(144) - 16(12)) = -(96 - 192) = -(-96) = 96 \ J$.Total work $W = 104 + 50 + 96 = 250 \ J$.

The components of a force acting on a particle vary according to the graphs shown. To reach point $B\, (8, 20, 0)$ from point $A\, (0, 5, 12)$,the particle moves on paths parallel to the $x$-axis,then the $y$-axis,and then the $z$-axis. The work done by this force is ............ $J$.

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