From the diagram to the applied force

Problem

A block of mass $m = 4{,}0\ \text{kg}$ moves with constant acceleration $a = 2{,}5\ \text{m/s}^2$ to the right on a horizontal plane with kinetic friction $\mu_d = 0{,}20$. Find the applied horizontal force.

Solution

We build the free-body diagram: acting on the block are the weight $\vv{P}$ (downward), the normal reaction $\vv{N}$ (upward), the applied force $\vv{F}$ (horizontal, to the right) and the kinetic friction $\vv{F}_a$ (horizontal, opposing the motion, so to the left).

Vertical direction. The block has no vertical acceleration, so: $N = mg$

The kinetic friction force is then: $F_a = \mu_d N = \mu_d\, mg$

Horizontal direction. We apply Newton’s second law, with the acceleration directed to the right: $F - F_a = ma$

We obtain the applied force: $F = ma + \mu_d\, mg$

Substituting the numerical values with $g = 9{,}81\ \text{m/s}^2$: $F = 4{,}0 \cdot 2{,}5 + 0{,}20 \cdot 4{,}0 \cdot 9{,}81 = 10{,}0 + 7{,}85$

$\ev{F \approx 17{,}8\ \text{N}}$

The applied force must do two things together: overcome friction ($7{,}85\ \text{N}$) and provide the block’s acceleration ($10{,}0\ \text{N}$).

Links

Topics: Dynamics · Friction Concepts: Newton’s second law · Kinetic friction Skills: Free-body diagram Objects: Block