Newton's Second Law Problems (F = ma) Step by Step

$$\Sigma \vec{F} = m\vec{a}$$

$\Sigma \vec{F}$ is the net force, the vector sum of every force on the object. Force is in newtons (N), mass in kg, and acceleration in $\text{m/s}^2$.

Because force is a vector, apply the law separately along each axis:

$$\Sigma F_x = ma_x \qquad \Sigma F_y = ma_y$$

The five step routine:

1. Draw a free-body diagram: a dot for the object and an arrow for every force acting on it.
2. Choose axes. Point one axis in the direction of acceleration if you know it.
3. Break angled forces into components.
4. Write $\Sigma F = ma$ for each axis. If there is no acceleration along an axis, that sum equals zero.
5. Solve for the unknown.

Force	Symbol	Direction	Size
Weight (gravity)	$W$ or $F_g$	Straight down	$mg$
Normal force	$N$	Perpendicular to the surface, pushing out	Found from $\Sigma F = ma$
Kinetic friction	$f_k$	Opposite the sliding motion	$\mu_k N$
Static friction	$f_s$	Opposite the motion it prevents	Up to $\mu_s N$
Tension	$T$	Along the rope, pulling away from the object	Found from $\Sigma F = ma$
Applied force	$F$	As given	As given

Only draw forces that act on the object. Do not draw "ma" as a force. It is the result of the forces, not one of them.

Use $g = 9.8 \text{ m/s}^2$ below.

Problem: A 10.0 kg box is pushed across the floor with a horizontal force of 50.0 N. The coefficient of kinetic friction is 0.30. Find the acceleration.

1. Forces: push 50.0 N to the right, friction to the left, weight down, normal force up.
2. y-axis (no vertical acceleration):

$$N - mg = 0 \quad\Rightarrow\quad N = (10.0)(9.8) = 98 \text{ N}$$

3. Friction: $f_k = \mu_k N = 0.30 \times 98 = 29.4$ N.
4. x-axis:

$$50.0 - 29.4 = 10.0a \quad\Rightarrow\quad a = \frac{20.6}{10.0} = 2.06 \text{ m/s}^2$$

Answer: about 2.1 $\text{m/s}^2$ to the right (two significant figures, set by 0.30).

An angled force changes the normal force, which changes friction.

Problem: A 20.0 kg sled is pulled with a 80.0 N force at $30.0^\circ$ above the horizontal. $\mu_k = 0.20$. Find the acceleration.

1. Components of the pull:
   • $F_x = 80.0\cos 30.0^\circ = 69.28$ N
   • $F_y = 80.0\sin 30.0^\circ = 40.0$ N (upward)
2. y-axis: the pull helps lift the sled, so the floor pushes up less.

$$N + 40.0 - (20.0)(9.8) = 0 \quad\Rightarrow\quad N = 196 - 40.0 = 156 \text{ N}$$

3. Friction: $f_k = 0.20 \times 156 = 31.2$ N.
4. x-axis:

$$69.28 - 31.2 = 20.0a \quad\Rightarrow\quad a = \frac{38.08}{20.0} = 1.90 \text{ m/s}^2$$

If you had used $N = mg = 196$ N, friction would be 39.2 N and the answer would be wrong.

On a ramp, tilt your axes: x along the slope and y perpendicular to it. Then only gravity needs to be split, using the incline angle $\theta$:

$$W_\parallel = mg\sin\theta \ (\text{down the slope}) \qquad W_\perp = mg\cos\theta \ (\text{into the slope})$$

Problem: A 5.0 kg block slides down a $30.0^\circ$ incline. $\mu_k = 0.20$. Find its acceleration.

1. Perpendicular axis: no acceleration into the ramp, so

$$N = mg\cos\theta = (5.0)(9.8)\cos 30.0^\circ = 42.4 \text{ N}$$

2. Friction (up the slope, opposing the slide): $f_k = 0.20 \times 42.4 = 8.49$ N.
3. Parallel component of gravity: $mg\sin\theta = (5.0)(9.8)(0.500) = 24.5$ N.
4. Along the slope (down is positive):

$$24.5 - 8.49 = 5.0a \quad\Rightarrow\quad a = \frac{16.0}{5.0} = 3.2 \text{ m/s}^2$$

Shortcut: the mass cancels, so $a = g(\sin\theta - \mu_k\cos\theta) = 9.8(0.500 - 0.20 \times 0.866) = 3.2 \text{ m/s}^2$.

Will it slide at all? A block at rest only starts sliding if $\tan\theta > \mu_s$. Here $\tan 30.0^\circ = 0.577$, so the block slides if $\mu_s$ is less than 0.577. With no friction, $a = g\sin\theta = 4.9 \text{ m/s}^2$.

Problem: A 3.0 kg block on a frictionless table is tied over a pulley to a 2.0 kg block hanging off the edge. Find the acceleration and the rope tension.

1. Treat the whole system first. The only unbalanced force driving the system is the hanging block's weight: $(2.0)(9.8) = 19.6$ N. The total mass is 5.0 kg.

$$a = \frac{19.6}{5.0} = 3.92 \text{ m/s}^2$$

2. Now one block alone to find tension. The only horizontal force on the table block is $T$:

$$T = m_1a = (3.0)(3.92) = 11.8 \text{ N}$$

3. Check with the hanging block: $m_2g - T = m_2a$ gives $19.6 - 11.8 = 7.8$ and $(2.0)(3.92) = 7.8$. It matches.

Notice the tension (11.8 N) is less than the hanging weight (19.6 N). If they were equal, the hanging block would not accelerate. Paste it into the VeilGPT solver to check your steps on harder pulley problems.