3000 N on a 1500 kg car
- Problem
- F = ma
- Solve for
- Acceleration
- Net force F
- 3000 N
- Mass m
- 1500 kg
- Show results in
- SI
- Acceleration
- 2 m/s²
Checked against: a = F/m = 3000/1500
Solve Newton's second law F = ma for force, mass or acceleration, or find acceleration, normal force and friction for a block on an inclined plane.
By AdminDrafted with AI writing toolsUpdated Checked examples: 6
A net force of 3,000 N on 1,500 kg gives an acceleration of 2 m/s² (0.2039 g).
Newton's second law says the net force on a body equals its mass times its acceleration, F = ma, so one newton gives 1 kg an acceleration of 1 m/s². Enter two of force, mass and acceleration to get the third. The inclined-plane mode splits the block's weight into mg sin θ down the slope and mg cos θ into it; the normal force balances the second part, and kinetic friction μk × N opposes the sliding.
The default, 3,000 N on a 1,500 kg car, gives 2 m/s², about 0.2 g. On the slope, a 10 kg block on a 30° incline with μk = 0.2 slides down at 3.20 m/s² against 16.99 N of friction.
Friction coefficients depend on both surfaces and their condition, so treat slope results as estimates. The block stays put while tan θ is at or below the static coefficient μs; leave μs blank and μk is used as that threshold.
Newton's second law says net force equals mass times acceleration: F = ma, so a = F/m. A 3,000 N net force on a 1,500 kg car gives 3,000 ÷ 1,500 = 2 m/s², about 0.2 g. With friction, subtract it from the push first: on level ground, a = (F − μk·mg)/m.
Opposing forces subtract. If a car's tires push forward with 3,500 N while drag and rolling resistance push back with 500 N, the net force is 3,000 N and the car accelerates at 2 m/s².
Weight is a separate force. The 1,500 kg car weighs 1,500 × 9.80665 = 14,710 N, but on level ground the road's normal force cancels it, so it leaves the horizontal acceleration unchanged.
One newton gives 1 kg an acceleration of 1 m/s². A pound-force is 4.4482216152605 N, the weight of 1 lb under standard gravity, so 100 lbf accelerating an object at 1 g means a mass of 100 lb, or 45.36 kg.
Friction is the coefficient of friction times the normal force: f = μN. On level ground N = mg, so sliding a 25 kg wooden box across a wooden floor with μk = 0.3 takes 0.3 × 25 × 9.80665 = 73.55 N at constant speed.
Push that box with 100 N and the net force is 100 − 73.55 = 26.45 N, so it accelerates at 26.45 ÷ 25 = 1.06 m/s².
Static friction is a limit, not a fixed value. It matches whatever force tries to start the motion, up to μs × N. Beyond that the object slips, and kinetic friction, μk × N, takes over.
Approximate coefficients from OpenStax University Physics, Table 6.1, static then kinetic:
Surface finish, moisture and temperature all move these numbers, so a coefficient measured on your own surfaces is more reliable.
On a slope at angle θ, split the weight mg into mg sin θ pulling down the slope and mg cos θ pressing into it. The normal force balances the second part, N = mg cos θ, so kinetic friction is μk mg cos θ.
For a 10 kg block on a 30° slope with μk = 0.2:
In one line, a = g(sin θ − μk cos θ). Mass cancels, so a heavier block slides no faster when friction is proportional to its weight.
A block at rest starts to slide when the down-slope pull beats the most static friction can supply: mg sin θ > μs mg cos θ, which simplifies to tan θ > μs. The critical angle is θc = arctan μs.
For wood on wood, μs = 0.5 gives θc = 26.57°. A 10 kg wooden block holds at 20°, where friction supplies just the 33.54 N needed. At 30° it slides, accelerating at 2.36 m/s² with μk = 0.3.
While the block holds, friction equals the down-slope pull, not μs N. On a 30° slope with μs = 0.7, the 10 kg block needs 49.03 N of friction, under the 59.45 N limit (HyperPhysics).
Default is 3,000 N on 1,500 kg. Slope is the 10 kg, 30°, μk = 0.2 block worked above. Static holds is the same block with μk = 0.5 and μs = 0.7, where friction cancels the pull and the net force is zero.
| Scenario | Acceleration (m/s²) | Net force (N) |
|---|---|---|
| Default | 2 | 3,000 |
| Slope | 3.20476 | 32.0476 |
| Static holds | 0 | 0 |
To turn a constant acceleration into distance and time, use the kinematics calculator. For the inward force on a curved path, use the circular motion calculator.
Checked against: a = F/m = 3000/1500
Checked against: 2 × 9.80665 = 19.6133
Checked against: Definition of the pound-force (NIST SP 811 B.8): 1 lbf = 0.45359237 kg × 9.80665 m/s²
Checked against: N = mg cos30° = 84.92808…, f = 0.2N, a = g(sin30° − 0.2 cos30°) = 3.2047634
One newton is the force that gives a 1 kg mass an acceleration of 1 m/s², so 1 N = 1 kg·m/s² in the SI. A 1 kg mass weighs 9.80665 N under standard gravity, and one pound-force is exactly 4.4482216152605 N (NIST SP 811). A 100 lbf push that accelerates an object at 1 g therefore means a mass of 100 lb, or 45.36 kg.
Friction equals the coefficient of friction times the normal force, f = μN. On level ground N = mg, so a 10 kg crate with μk = 0.2 needs a 19.61 N push to keep sliding at constant speed. Static friction is a limit, not a fixed value: it matches the applied force up to μs × N, and the object starts to move once that limit is passed.
It starts to slide once the slope angle exceeds arctan μs, the point where the pull down the slope, mg sin θ, overtakes the most static friction can supply, μs mg cos θ. With μs = 0.7 that angle is 34.99°; with μs = 0.2 it is 11.31°. Tilting a surface until an object slips and reading the angle is a standard way to measure μs.
OpenStax University Physics Table 6.1 gives approximate values, static then kinetic: rubber on dry concrete 1.0 and 0.7, wood on wood 0.5 and 0.3, ice on ice 0.1 and 0.03. Real values shift with surface finish, moisture and temperature, so a measured coefficient beats a table value. The kinetic coefficient is lower than the static one for each of these pairs.
No. Mass cancels in a = g(sin θ − μk cos θ), because both the pull down the slope and the friction are proportional to the weight. A 5 kg block and a 50 kg block on a frictionless 45° slope both accelerate at 6.93 m/s². Differences seen in practice come from air resistance and from friction coefficients that change with load or speed.
This calculator includes 6 worked examples with independently sourced answers. They run in the test suite, and you can run them here too.