Judging from your answer... it seems that the weight and the y-component of the applied force are in the same direction; vertically downwards.

I think your answer is wrong because you've put some negative signs on your components that shouldn't be there. Here's a full solution to explain.

Assuming that the applied force *is* pointing downwards by 25 degrees from the horizontal, instead of upwards, we have the following forces labelled on the attached free-body diagram:

1. Weight = \(\displaystyle mg\)

2. Reaction force, \(\displaystyle R\)

3. Friction force, \(\displaystyle F_r \le \mu R\),

4. Applied force = \(\displaystyle F_{app}\) = 100 N

Let's consider the y-direction. Since we know that the object accelerates along the surface, we know that the reaction force must balance the weight and the applied force. Therefore, according to Newton's first law, the reaction force balances the sum of the weight and y-component of the applied force:

\(\displaystyle R = mg + F_{app} sin(25)\)

Let's now consider the x-direction. In this direction we know the object accelerates along the surface. Therefore, we can solve for Newton's second law along the x-direction. The total force in the x-direction is the difference between the x-component of the applied force and the friction force, which always resists motion:

\(\displaystyle F_{total,x} = ma = F_{app} cos(25) - F_r\)

Because the object is moving, we know the limiting friction force has been overcome, so we know that the friction force relates to the reaction force exactly using:

\(\displaystyle F_r = \mu R\)

We can now substitute these quantities in the second equation and solve for acceleration:

\(\displaystyle ma = F_{app} cos(25) - \mu R\)

\(\displaystyle = F_{app} cos(25) - \mu (mg + F_{app} sin(25))\)

\(\displaystyle a = \frac{F_{app} cos(25) - \mu (mg + F_{app} sin(25))}{m}\)

Substituting values:

\(\displaystyle a = \frac{100 \times cos(25) - 0.12 \times (15 \times 9.81 + 100 \times sin(25))}{15} = 4.527 m/s^2\)