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)

Solution & Explanation

Let us resolve the applied force $$F$$ acting on the block of mass $$m$$ into its orthogonal vector components. The force is applied at an angle $$\theta$$ relative to the horizontal surface:

• Horizontal component ($$F_x$$): Drives the block forward along the floor.

  $$F_x = F \cos\theta$$

• Vertical component ($$F_y$$): Pushes the block downward, directly increasing the contact pressure with the floor.

  $$F_y = F \sin\theta$$

To find the normal reaction force ($$N$$) exerted by the ground, we set up the static equilibrium equation along the vertical axis where there is no net motion:

$$\Sigma F_{\text{vertical}} = 0 \implies N - mg - F_y = 0$$

$$N = mg + F \sin\theta$$

The kinetic friction force ($$f_K$$) opposing the forward sliding motion of the block depends directly on this normal reaction force:

$$f_K = \mu_K N$$

$$f_K = \mu_K (mg + F \sin\theta)$$

Applying Newton's Second Law along the horizontal axis of motion yields the net force equation:

$$F_{\text{net}} = F_x - f_K = ma$$

$$F \cos\theta - \mu_K (mg + F \sin\theta) = ma$$

To explicitly solve for the block's acceleration ($$a$$), divide the entire expression by the mass $$m$$:

$$a = \frac{F \cos\theta - \mu_K (mg + F \sin\theta)}{m}$$

$$a = \frac{F}{m} \cos\theta - \mu_K \left(g + \frac{F}{m} \sin\theta\right)$$

Concept Check: Because the external force pushes downward at an angle, it actively presses the block tighter into the floor. This structural alignment augments the standard normal force from $$mg$$ to $$mg + F\sin\theta$$, ramping up the kinetic friction drag and reducing the net forward acceleration to exactly $$\frac{F}{m} \cos\theta - \mu_K \left(g + \frac{F}{m} \sin\theta\right)$$.

Correct Option Choice: Option C $$\left(\frac{F}{m} \cos\theta - \mu_K \left(g + \frac{F}{m} \sin\theta\right)\right)$$