Use this free online force converter to change joules per meter into femtonewtons instantly. Type in the joules per meter value, and the equivalent femtonewtons is calculated for you in real time.
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Enter your inputs, and the result is calculated in real-time.
Joules per Meter
Femtonewtons
How to use this Joules per Meter to Femtonewtons Converter 🤔
Follow these steps to convert given Joules per Meter value from Joules per Meter units to Femtonewtons units.
Enter the input Joules per Meter value in the text field.
The given Joules per Meter is converted to Femtonewtons in realtime ⌚ using the formula, and displayed under the Femtonewtons label.
You may copy the resulting Femtonewtons value using the Copy button.
Formula
To convert given force from Joules per Meter to Femtonewtons, use the following formula.
Femtonewtons = Joules per Meter * 1e+15
Calculation
Calculation will be done after you enter a valid input.
Joules per Meter
Joule per meter (J/m) is a unit that represents energy per unit length. It is often used in the context of surface tension, where it describes the energy required to increase the surface area of a liquid.
Femtonewtons
A femtonewton (fN) is one-quadrillionth of a newton. It is used in experimental physics and nanotechnology to measure forces at the atomic scale, such as the forces involved in the interaction between nanoparticles.
{
"conversion": "joule-per-meter-femtonewton",
"x_slug": "joule-per-meter",
"y_slug": "femtonewton",
"x": "J/m",
"y": "fN",
"x_desc": "Joules per Meter",
"y_desc": "Femtonewtons",
"category": "Force",
"symbol": "m",
"formula": "x * 1e+15",
"examples": "<div class=\"example\">\n <div class=\"example_head\"><span class=\"example_n\">1</span>\n <h3 class=\"question\">Consider a force applied over a distance of 500 meters using an energy of 1,000 joules per meter.<br>Convert this force from joule-per-meter to Femtonewtons.</h3></div>\n <h4 class=\"answer\">Answer:</h4>\n <p><strong>Given:</strong></p>\n <p>The force of force over distance in joules per meter is:</p>\n <p class=\"step\"><span>Force<sub>(Joules per Meter)</sub></span> = 1000</p>\n <p><strong>Formula:</strong></p>\n <p>The formula to convert force from joules per meter to femtonewtons is:</p>\n <p class=\"formula step\"><span>Force<sub>(Femtonewtons)</sub></span> = <span>Force<sub>(Joules per Meter)</sub></span> × 1e+15</p>\n <p><strong>Substitution:</strong></p>\n <p>Substitute given weight of force over distance, <strong>Force<sub>(Joules per Meter)</sub> = 1000</strong> in the above formula.</p>\n <p class=\"step\"><span>Force<sub>(Femtonewtons)</sub></span> = <span>1000</span> × 1e+15</p>\n <p class=\"step\"><span>Force<sub>(Femtonewtons)</sub></span> = 1000000000000000000</p>\n <p><strong>Final Answer:</strong></p>\n <p>Therefore, <strong>1000 J/m</strong> is equal to <strong>1000000000000000000 fN</strong>.</p>\n <p>The force of force over distance is <strong>1000000000000000000 fN</strong>, in femtonewtons.</p>\n </div>\n <div class=\"example\">\n <div class=\"example_head\"><span class=\"example_n\">2</span>\n <h3 class=\"question\">Consider an engine producing 2,000 joules per meter to move a vehicle.<br>Convert this force from joule-per-meter to Femtonewtons.</h3></div>\n <h4 class=\"answer\">Answer:</h4>\n <p><strong>Given:</strong></p>\n <p>The force of engine to move vehicle in joules per meter is:</p>\n <p class=\"step\"><span>Force<sub>(Joules per Meter)</sub></span> = 2000</p>\n <p><strong>Formula:</strong></p>\n <p>The formula to convert force from joules per meter to femtonewtons is:</p>\n <p class=\"formula step\"><span>Force<sub>(Femtonewtons)</sub></span> = <span>Force<sub>(Joules per Meter)</sub></span> × 1e+15</p>\n <p><strong>Substitution:</strong></p>\n <p>Substitute given weight of engine to move vehicle, <strong>Force<sub>(Joules per Meter)</sub> = 2000</strong> in the above formula.</p>\n <p class=\"step\"><span>Force<sub>(Femtonewtons)</sub></span> = <span>2000</span> × 1e+15</p>\n <p class=\"step\"><span>Force<sub>(Femtonewtons)</sub></span> = 2000000000000000000</p>\n <p><strong>Final Answer:</strong></p>\n <p>Therefore, <strong>2000 J/m</strong> is equal to <strong>2000000000000000000 fN</strong>.</p>\n <p>The force of engine to move vehicle is <strong>2000000000000000000 fN</strong>, in femtonewtons.</p>\n </div>\n ",
"units": [
[
"newton",
"Newtons",
"N"
],
[
"kilonewton",
"Kilonewtons",
"kN"
],
[
"gram-force",
"Gram-Force",
"gf"
],
[
"kilogram-force",
"Kilogram-Force",
"kgf"
],
[
"ton-force",
"Metric Ton-Force",
"tf"
],
[
"exanewton",
"Exanewtons",
"EN"
],
[
"petanewton",
"Petanewtons",
"PT"
],
[
"teranewton",
"Teranewtons",
"TN"
],
[
"giganewton",
"Giganewtons",
"GN"
],
[
"meganewton",
"Meganewtons",
"MN"
],
[
"hectonewton",
"Hectonewtons",
"hN"
],
[
"dekanewton",
"Dekanewtons",
"daN"
],
[
"decinewton",
"Decinewtons",
"dN"
],
[
"centinewton",
"Centinewtons",
"cN"
],
[
"millinewton",
"Millinewtons",
"mN"
],
[
"micronewton",
"Micronewtons",
"µN"
],
[
"nanonewton",
"Nanonewtons",
"nN"
],
[
"piconewton",
"Piconewtons",
"pN"
],
[
"femtonewton",
"Femtonewtons",
"fN"
],
[
"attonewton",
"Attonewtons",
"aN"
],
[
"dyne",
"Dynes",
"dyn"
],
[
"joule-per-meter",
"Joules per Meter",
"J/m"
],
[
"joule-per-centimeter",
"Joules per Centimeter",
"J/cm"
],
[
"ton-force-short",
"Short Ton-Force",
"short tonf"
],
[
"to-force-long",
"Long Ton-Force (UK)",
"tonf (UK)"
],
[
"kip-force",
"Kip-Force",
"kipf"
],
[
"kilopound-force",
"Kilopound-Force",
"kipf"
],
[
"pound-force",
"Pound-Force",
"lbf"
],
[
"ounce-force",
"Ounce-Force",
"ozf"
],
[
"poundal",
"Poundals",
"pdl"
],
[
"pound-foot-per-square-second",
"Pound Foot per Square Second",
"lbf·ft/s²"
],
[
"pond",
"Ponds",
"p"
],
[
"kilopond",
"Kiloponds",
"kp"
]
],
"y_long_desc": "A femtonewton (fN) is one-quadrillionth of a newton. It is used in experimental physics and nanotechnology to measure forces at the atomic scale, such as the forces involved in the interaction between nanoparticles.",
"x_long_desc": "Joule per meter (J/m) is a unit that represents energy per unit length. It is often used in the context of surface tension, where it describes the energy required to increase the surface area of a liquid."
}
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