Answer:
1.77 m/s^2
Explanation:
There are two forces acting on the car along the direction parallel to the incline:
- The driving force of 10,000 N, which pushes forward
- The component of the weigth of the car parallel to the incline, which pulls backward
The component of the weight of the car parallel to the incline is:
[tex]W_p = mg sin \theta=(1500 kg)(9.8 m/s^2)( sin 30^{\circ})=7350 N[/tex]
So now we can apply Newton's second law to find the acceleration of the car:
[tex]F-W_p = ma\\a=\frac{F-W_p}{m}=\frac{10000 N-7350 N}{1500 kg}=1.77 m/s^2[/tex]
