Answer :
We begin by resolving each force into its horizontal (x) and vertical (y) components.
1. For the first force of [tex]$124 \, \text{N}$[/tex] acting at an angle of [tex]$29.6^\circ$[/tex], the components are calculated as follows:
[tex]$$F_{1x} = 124 \cos(29.6^\circ) \approx 107.82 \, \text{N},$$[/tex]
[tex]$$F_{1y} = 124 \sin(29.6^\circ) \approx 61.25 \, \text{N}.$$[/tex]
2. For the second force of [tex]$187 \, \text{N}$[/tex] acting at an angle of [tex]$79.4^\circ$[/tex], the components are:
[tex]$$F_{2x} = 187 \cos(79.4^\circ) \approx 34.40 \, \text{N},$$[/tex]
[tex]$$F_{2y} = 187 \sin(79.4^\circ) \approx 183.81 \, \text{N}.$$[/tex]
3. Next, we find the net force in each direction by summing the corresponding components:
[tex]$$F_{\text{net},x} = F_{1x} + F_{2x} \approx 107.82 + 34.40 = 142.22 \, \text{N},$$[/tex]
[tex]$$F_{\text{net},y} = F_{1y} + F_{2y} \approx 61.25 + 183.81 = 245.06 \, \text{N}.$$[/tex]
4. The magnitude of the net force is found using the Pythagorean theorem:
[tex]$$F_{\text{net}} = \sqrt{F_{\text{net},x}^2 + F_{\text{net},y}^2} \approx \sqrt{(142.22)^2 + (245.06)^2} \approx 283.33 \, \text{N}.$$[/tex]
5. Finally, applying Newton's second law, the acceleration [tex]$a$[/tex] of the block (mass [tex]$m = 305 \, \text{kg}$[/tex]) is:
[tex]$$a = \frac{F_{\text{net}}}{m} \approx \frac{283.33}{305} \approx 0.93 \, \text{m/s}^2.$$[/tex]
Thus, the acceleration of the block is approximately
[tex]$$\boxed{0.93 \, \text{m/s}^2}.$$[/tex]
1. For the first force of [tex]$124 \, \text{N}$[/tex] acting at an angle of [tex]$29.6^\circ$[/tex], the components are calculated as follows:
[tex]$$F_{1x} = 124 \cos(29.6^\circ) \approx 107.82 \, \text{N},$$[/tex]
[tex]$$F_{1y} = 124 \sin(29.6^\circ) \approx 61.25 \, \text{N}.$$[/tex]
2. For the second force of [tex]$187 \, \text{N}$[/tex] acting at an angle of [tex]$79.4^\circ$[/tex], the components are:
[tex]$$F_{2x} = 187 \cos(79.4^\circ) \approx 34.40 \, \text{N},$$[/tex]
[tex]$$F_{2y} = 187 \sin(79.4^\circ) \approx 183.81 \, \text{N}.$$[/tex]
3. Next, we find the net force in each direction by summing the corresponding components:
[tex]$$F_{\text{net},x} = F_{1x} + F_{2x} \approx 107.82 + 34.40 = 142.22 \, \text{N},$$[/tex]
[tex]$$F_{\text{net},y} = F_{1y} + F_{2y} \approx 61.25 + 183.81 = 245.06 \, \text{N}.$$[/tex]
4. The magnitude of the net force is found using the Pythagorean theorem:
[tex]$$F_{\text{net}} = \sqrt{F_{\text{net},x}^2 + F_{\text{net},y}^2} \approx \sqrt{(142.22)^2 + (245.06)^2} \approx 283.33 \, \text{N}.$$[/tex]
5. Finally, applying Newton's second law, the acceleration [tex]$a$[/tex] of the block (mass [tex]$m = 305 \, \text{kg}$[/tex]) is:
[tex]$$a = \frac{F_{\text{net}}}{m} \approx \frac{283.33}{305} \approx 0.93 \, \text{m/s}^2.$$[/tex]
Thus, the acceleration of the block is approximately
[tex]$$\boxed{0.93 \, \text{m/s}^2}.$$[/tex]
