Use this free online force converter to change kilogram-force into joules per meter instantly. Type in the kilogram-force value, and the equivalent joules per meter is calculated for you in real time.
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Kilogram-Force
Joules per Meter
How to use this Kilogram-Force to Joules per Meter Converter 🤔
Follow these steps to convert given Kilogram-Force value from Kilogram-Force units to Joules per Meter units.
Enter the input Kilogram-Force value in the text field.
The given Kilogram-Force is converted to Joules per Meter in realtime ⌚ using the formula, and displayed under the Joules per Meter label.
You may copy the resulting Joules per Meter value using the Copy button.
Formula
To convert given force from Kilogram-Force to Joules per Meter, use the following formula.
Joules per Meter = Kilogram-Force * 9.806650000000271
Calculation
Calculation will be done after you enter a valid input.
Kilogram-Force
Kilogram-force (kgf) is the force exerted by the weight of one kilogram under standard gravity. It is often used in engineering to describe the force of everyday objects and loads, such as the force exerted by a heavy tool or piece of equipment. It is a non-SI unit but still finds practical use in many fields.
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.
{
"conversion": "kilogram-force-joule-per-meter",
"x_slug": "kilogram-force",
"y_slug": "joule-per-meter",
"x": "kgf",
"y": "J/m",
"x_desc": "Kilogram-Force",
"y_desc": "Joules per Meter",
"category": "Force",
"symbol": "m",
"formula": "x * 9.806650000000271",
"examples": "<div class=\"example\">\n <div class=\"example_head\"><span class=\"example_n\">1</span>\n <h3 class=\"question\">Consider a person applying 70 kilogram-force to lift a weight in the gym.<br>Convert this force from kilogram-force to Joules per Meter.</h3></div>\n <h4 class=\"answer\">Answer:</h4>\n <p><strong>Given:</strong></p>\n <p>The force of lifting the weight in kilogram-force is:</p>\n <p class=\"step\"><span>Force<sub>(Kilogram-Force)</sub></span> = 70</p>\n <p><strong>Formula:</strong></p>\n <p>The formula to convert force from kilogram-force to joules per meter is:</p>\n <p class=\"formula step\"><span>Force<sub>(Joules per Meter)</sub></span> = <span>Force<sub>(Kilogram-Force)</sub></span> × 9.806650000000271</p>\n <p><strong>Substitution:</strong></p>\n <p>Substitute given weight of lifting the weight, <strong>Force<sub>(Kilogram-Force)</sub> = 70</strong> in the above formula.</p>\n <p class=\"step\"><span>Force<sub>(Joules per Meter)</sub></span> = <span>70</span> × 9.806650000000271</p>\n <p class=\"step\"><span>Force<sub>(Joules per Meter)</sub></span> = 686.4655</p>\n <p><strong>Final Answer:</strong></p>\n <p>Therefore, <strong>70 kgf</strong> is equal to <strong>686.4655 J/m</strong>.</p>\n <p>The force of lifting the weight is <strong>686.4655 J/m</strong>, in joules per meter.</p>\n </div>\n <div class=\"example\">\n <div class=\"example_head\"><span class=\"example_n\">2</span>\n <h3 class=\"question\">Consider a shopping bag that exerts 5 kilogram-force.<br>Convert this force from kilogram-force to Joules per Meter.</h3></div>\n <h4 class=\"answer\">Answer:</h4>\n <p><strong>Given:</strong></p>\n <p>The force of shopping bag in kilogram-force is:</p>\n <p class=\"step\"><span>Force<sub>(Kilogram-Force)</sub></span> = 5</p>\n <p><strong>Formula:</strong></p>\n <p>The formula to convert force from kilogram-force to joules per meter is:</p>\n <p class=\"formula step\"><span>Force<sub>(Joules per Meter)</sub></span> = <span>Force<sub>(Kilogram-Force)</sub></span> × 9.806650000000271</p>\n <p><strong>Substitution:</strong></p>\n <p>Substitute given weight of shopping bag, <strong>Force<sub>(Kilogram-Force)</sub> = 5</strong> in the above formula.</p>\n <p class=\"step\"><span>Force<sub>(Joules per Meter)</sub></span> = <span>5</span> × 9.806650000000271</p>\n <p class=\"step\"><span>Force<sub>(Joules per Meter)</sub></span> = 49.0333</p>\n <p><strong>Final Answer:</strong></p>\n <p>Therefore, <strong>5 kgf</strong> is equal to <strong>49.0333 J/m</strong>.</p>\n <p>The force of shopping bag is <strong>49.0333 J/m</strong>, in joules per meter.</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"
]
],
"x_long_desc": "Kilogram-force (kgf) is the force exerted by the weight of one kilogram under standard gravity. It is often used in engineering to describe the force of everyday objects and loads, such as the force exerted by a heavy tool or piece of equipment. It is a non-SI unit but still finds practical use in many fields.",
"y_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."
}
ConvertOnline