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Answer :
We start by breaking each force into its horizontal and vertical components.
1. For the first force of [tex]$124\,N$[/tex] at an angle of [tex]$29.6^\circ$[/tex], the components are:
[tex]$$
F_{1x} = 124 \cos(29.6^\circ) \approx 107.82\,N
$$[/tex]
[tex]$$
F_{1y} = 124 \sin(29.6^\circ) \approx 61.25\,N
$$[/tex]
2. For the second force of [tex]$187\,N$[/tex] at an angle of [tex]$79.4^\circ$[/tex], the components are:
[tex]$$
F_{2x} = 187 \cos(79.4^\circ) \approx 34.40\,N
$$[/tex]
[tex]$$
F_{2y} = 187 \sin(79.4^\circ) \approx 183.81\,N
$$[/tex]
3. Next, we find the net force components by summing the corresponding components:
[tex]$$
F_{\text{net},x} = F_{1x} + F_{2x} \approx 107.82\,N + 34.40\,N = 142.22\,N
$$[/tex]
[tex]$$
F_{\text{net},y} = F_{1y} + F_{2y} \approx 61.25\,N + 183.81\,N = 245.06\,N
$$[/tex]
4. The magnitude of the net force is found using the Pythagorean theorem:
[tex]$$
F_{\text{net}} = \sqrt{F_{\text{net},x}^2 + F_{\text{net},y}^2} \approx \sqrt{(142.22)^2 + (245.06)^2} \approx 283.34\,N
$$[/tex]
5. Finally, using Newton's second law [tex]$a = \frac{F}{m}$[/tex] with a mass of [tex]$305\,kg$[/tex], the acceleration is:
[tex]$$
a = \frac{F_{\text{net}}}{m} \approx \frac{283.34\,N}{305\,kg} \approx 0.929\,\frac{m}{s^2}
$$[/tex]
Thus, the acceleration of the block is approximately
[tex]$$
\boxed{0.93\,\frac{m}{s^2}}.
$$[/tex]
1. For the first force of [tex]$124\,N$[/tex] at an angle of [tex]$29.6^\circ$[/tex], the components are:
[tex]$$
F_{1x} = 124 \cos(29.6^\circ) \approx 107.82\,N
$$[/tex]
[tex]$$
F_{1y} = 124 \sin(29.6^\circ) \approx 61.25\,N
$$[/tex]
2. For the second force of [tex]$187\,N$[/tex] at an angle of [tex]$79.4^\circ$[/tex], the components are:
[tex]$$
F_{2x} = 187 \cos(79.4^\circ) \approx 34.40\,N
$$[/tex]
[tex]$$
F_{2y} = 187 \sin(79.4^\circ) \approx 183.81\,N
$$[/tex]
3. Next, we find the net force components by summing the corresponding components:
[tex]$$
F_{\text{net},x} = F_{1x} + F_{2x} \approx 107.82\,N + 34.40\,N = 142.22\,N
$$[/tex]
[tex]$$
F_{\text{net},y} = F_{1y} + F_{2y} \approx 61.25\,N + 183.81\,N = 245.06\,N
$$[/tex]
4. The magnitude of the net force is found using the Pythagorean theorem:
[tex]$$
F_{\text{net}} = \sqrt{F_{\text{net},x}^2 + F_{\text{net},y}^2} \approx \sqrt{(142.22)^2 + (245.06)^2} \approx 283.34\,N
$$[/tex]
5. Finally, using Newton's second law [tex]$a = \frac{F}{m}$[/tex] with a mass of [tex]$305\,kg$[/tex], the acceleration is:
[tex]$$
a = \frac{F_{\text{net}}}{m} \approx \frac{283.34\,N}{305\,kg} \approx 0.929\,\frac{m}{s^2}
$$[/tex]
Thus, the acceleration of the block is approximately
[tex]$$
\boxed{0.93\,\frac{m}{s^2}}.
$$[/tex]
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