A block of mass $$m$$ slides along a floor while a force of magnitude $$F$$ is applied to it at an angle $$\theta$$ as shown in figure. The coefficient of kinetic friction is $$\mu_K$$. Then, the block's acceleration $$a$$ is given by: ($$g$$ is acceleration due to gravity)

We need to find the expression for the acceleration $$a$$ of a block of mass $$m$$ sliding along a floor under the action of a force $$F$$ applied at an angle $$\theta$$.

First, we resolve the applied force $$F$$ into two rectangular components:

• The horizontal component pushing the block forward: $$F_x = F \cos\theta$$.

• The vertical component pushing the block downward into the floor: $$F_y = F \sin\theta$$.

Next, we consider the forces acting along the vertical direction to find the normal reaction force $$N$$. The downward forces (gravity $$mg$$ and the vertical force component $$F \sin\theta$$) must balance the upward normal force: $$N = mg + F \sin\theta$$.

The kinetic friction force $$f_K$$ opposing the forward motion is given by the formula: $$f_K = \mu_K N = \mu_K (mg + F \sin\theta)$$.

Now, we write the net force equation along the horizontal direction of motion: $$F_{net} = F \cos\theta - f_K = m a$$.

Substituting the expression for $$f_K$$ into the equation yields: $$F \cos\theta - \mu_K (mg + F \sin\theta) = m a$$.

Dividing the entire equation by the mass $$m$$ to solve for acceleration $$a$$ gives: $$a = \frac{F}{m} \cos\theta - \mu_K \left(g + \frac{F}{m} \sin\theta\right)$$.

Therefore, the correct answer is Option C: $$\frac{F}{m} \cos\theta - \mu_K \left(g + \frac{F}{m} \sin\theta\right)$$.