Newton's Second Law Problems With Solutions: Net Force to Acceleration

Newton's second law says the net force on an object equals its mass times its acceleration: $\sum F=ma$. In a typical problem, draw the forces acting on the chosen object, pick a positive direction, add force components with their signs, and only then divide by the mass. A 4 kg object with a 20 N net force accelerates at 5 m/s².

The common trap is using just one force when several act. A $50$ N push opposed by $18$ N of friction produces $32$ N of net force, not $50$ N.

A repeatable four-step method

1. Choose the object or system. Only draw forces acting on that system. An object does not exert a force on itself.
2. Draw a free-body diagram. Label weight, normal force, tension, applied force and friction only when they really act. Our free-body-diagram guide goes deeper on this step.
3. Choose axes and write a signed net-force equation for each relevant axis. For a horizontal push, right might be positive. For an incline, an axis parallel to the slope is often easier.
4. Solve and check. Use $1\text{ N}=1\text{ kg}\cdot\text{m/s}^2$. Acceleration should point in the net-force direction.

The second law uses net force, which is a vector sum. Forces at right angles cannot simply be added as ordinary positive numbers. OpenStax's treatment of Newton's law and other forces shows how axes make those components manageable.

Example 1: one horizontal net force

Problem: A $4$ kg cart has a net horizontal force of $20$ N to the right. Find its acceleration.

Force picture: The problem already gives the net force. A full diagram of individual forces cannot be inferred from that information, but the resulting force arrow points right.

Answer: $5\text{ m/s}^2$ to the right. The units check: N/kg is $\text{m/s}^2$. Do not assume the cart's velocity is $5\text{ m/s}$; acceleration and velocity are different quantities.

Example 2: a push opposed by friction

Problem: An $8$ kg box moves on a level floor. A $50$ N force pushes it right, and kinetic friction is $18$ N left. Find its acceleration. Assume no vertical acceleration.

Free-body diagram, with the box as the system:

              normal force
                   ↑
friction 18 N ← [ box ] → push 50 N
                   ↓
                  weight

The normal force balances weight vertically. Horizontally, right is positive:

Answer: $4\text{ m/s}^2$ right. Friction reduces the acceleration but does not reverse it because the push is larger.

Example 3: tension in an accelerating elevator

Problem: A $3$ kg load hangs from a cable in an elevator accelerating upward at $2\text{ m/s}^2$. What is the cable tension? Use $g=9.8\text{ m/s}^2$ and ignore air resistance.

Force picture: Tension $T$ acts up on the load; weight $mg$ acts down. Choose up as positive. Weight is $3(9.8)=29.4$ N.

Answer: $35.4$ N. It is greater than the $29.4$ N weight, as it must be for upward acceleration. If the elevator moved upward at constant speed, acceleration would be zero and tension would equal weight.

Example 4: a block sliding down a rough incline

Problem: A $5$ kg block is sliding down a $30^\circ$ slope. The coefficient of kinetic friction is $\mu_k=0.20$. Find the acceleration down the slope. Use $g=9.8\text{ m/s}^2$.

Force picture: Weight $mg$ points vertically down; the normal force $N$ points perpendicular to the slope; kinetic friction $f_k$ points up the slope because it opposes the sliding. Choose downhill as the positive slope direction. Resolve weight into $mg\sin30^\circ$ downhill and $mg\cos30^\circ$ into the slope.

There is no acceleration perpendicular to the slope, so

Kinetic friction is $f_k=\mu_kN\approx0.20(42.44)=8.49$ N. The downhill weight component is $mg\sin30^\circ=24.5$ N. Therefore,

Check: Without friction, the acceleration would be $g\sin30^\circ=4.9\text{ m/s}^2$ downhill. Friction makes $3.20$ smaller, which is sensible. This calculation assumes the block is already sliding, so kinetic rather than static friction applies.

What to check before accepting an answer

If the question continues from force to distance or stopping time, use the acceleration from $\sum F=ma$ in the kinematics equation guide. The Physics Solver can help check a pictured setup, and physics resources collects related tools.

Practice answers