Answer :
Answer: [tex]25.08\ m/s^2[/tex]
Explanation:
Given
mass of wagon m=5 kg
elevation
[tex]\theta =30.9^{\circ}\\F=150\ N[/tex]
Sin component of weight will oppose the applied force therefore we can write
[tex]F-W\sin \theta=ma\\where\\W=weight(mg)\\a=acceleration[/tex]
[tex]150-5\times 9.8\times \sin (30.09)^{\circ}=5\times a\\125.43=5a\\a=25.08\ m/s^2[/tex]