Difference between revisions of "Newton's Second Law"
(Created page with "==Key Stage 4 Foundation== ===Meaning=== '''Newton's Second Law''' states that "Force = Mass x Acceleration, which means the acceleration of an object is Directly Pr...") |
|||
| (8 intermediate revisions by 2 users not shown) | |||
| Line 11: | Line 11: | ||
<math>F=ma</math> | <math>F=ma</math> | ||
| − | Where | + | Where |
| − | F = The [[Resultant Force]] on the [[object]]. | + | <math>F</math> = The [[Resultant Force]] on the [[object]]. |
| − | m = The [[mass]] of the [[object]]. | + | <math>m</math> = The [[mass]] of the [[object]]. |
| − | a = The [[acceleration]] of the [[object]]. | + | <math>a</math> = The [[acceleration]] of the [[object]]. |
===Example Calculations=== | ===Example Calculations=== | ||
| − | ====Finding the Force given Mass and Acceleration=== | + | ====Finding the Force given Mass and Acceleration==== |
{| class="wikitable" | {| class="wikitable" | ||
| − | | style="height:20px; width: | + | | style="height:20px; width:300px; text-align:center;" |A 2.3kg [[object]] [[accelerate]]s at a rate of 8.8m/s/s. Calculate the [[Resultant Force|resultant force]] acting on the [[object]] correct to two [[Significant Figures|significant figures]]. |
| − | | style="height:20px; width: | + | | style="height:20px; width:300px; text-align:center;" |A 5.5x10<sup>4</sup>kg rocket [[accelerate]]s at a rate of 61m/s/s. Calculate the [[Resultant Force|resultant force]] acting on the rocket correct to two [[Significant Figures|significant figures]]. |
|- | |- | ||
| − | | style="height:20px; width: | + | | style="height:20px; width:300px; text-align:left;" |'''1. State the known quantities''' |
m = 2.3kg | m = 2.3kg | ||
a = 8.8m/s/s | a = 8.8m/s/s | ||
| − | | style="height:20px; width: | + | | style="height:20px; width:300px; text-align:left;" |'''1. State the known quantities''' |
m = 5.5x10<sup>4</sup>kg | m = 5.5x10<sup>4</sup>kg | ||
| Line 36: | Line 36: | ||
a = 61m/s/s | a = 61m/s/s | ||
|- | |- | ||
| − | | style="height:20px; width: | + | | style="height:20px; width:300px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers into the [[equation]] and [[Solve (Maths)|solve]].''' |
<math>F=ma</math> | <math>F=ma</math> | ||
| Line 45: | Line 45: | ||
<math>F \approx 20N</math> | <math>F \approx 20N</math> | ||
| − | | style="height:20px; width: | + | | style="height:20px; width:300px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers into the [[equation]] and [[Solve (Maths)|solve]].''' |
<math>F=ma</math> | <math>F=ma</math> | ||
| Line 56: | Line 56: | ||
|} | |} | ||
| − | ====Finding the Acceleration given Mass and Force=== | + | ====Finding the Acceleration given Mass and Force==== |
{| class="wikitable" | {| class="wikitable" | ||
| − | | style="height:20px; width: | + | | style="height:20px; width:300px; text-align:center;" |A 7kg [[object]] is subjected to a [[Resultant Force|resultant force]] of 53N. Calculate the [[acceleration]] of the [[object]] correct to two [[Significant Figures|significant figures]]. |
| − | | style="height:20px; width: | + | | style="height:20px; width:300px; text-align:center;" |A 160g snooker ball experiences a [[Resultant Force|resultant force]] of 12N. Calculate the [[acceleration]] of the snooker ball correct to two [[Significant Figures|significant figures]]. |
|- | |- | ||
| − | | style="height:20px; width: | + | | style="height:20px; width:300px; text-align:left;" |'''1. State the known quantities''' |
m = 7kg | m = 7kg | ||
F = 53N | F = 53N | ||
| − | | style="height:20px; width: | + | | style="height:20px; width:300px; text-align:left;" |'''1. State the known quantities''' |
m = 160g = 0.16kg | m = 160g = 0.16kg | ||
| Line 72: | Line 72: | ||
F = 12N | F = 12N | ||
|- | |- | ||
| − | | style="height:20px; width: | + | | style="height:20px; width:300px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers and [[Evaluate (Maths)|evaluate]].''' |
<math>F=ma</math> | <math>F=ma</math> | ||
<math>53=7a</math> | <math>53=7a</math> | ||
| − | | style="height:20px; width: | + | | style="height:20px; width:300px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers and [[Evaluate (Maths)|evaluate]].''' |
<math>F=ma</math> | <math>F=ma</math> | ||
| Line 83: | Line 83: | ||
<math>12=0.16a</math> | <math>12=0.16a</math> | ||
|- | |- | ||
| − | | style="height:20px; width: | + | | style="height:20px; width:300px; text-align:left;" |'''3. [[Rearrange (Maths)|Rearrange]] the equation and [[Solve (Maths)|solve]].''' |
<math>a = \frac{53}{7}</math> | <math>a = \frac{53}{7}</math> | ||
| Line 89: | Line 89: | ||
<math>a \approx 7.6m/s/s</math> | <math>a \approx 7.6m/s/s</math> | ||
| − | | style="height:20px; width: | + | | style="height:20px; width:300px; text-align:left;" |'''3. [[Rearrange (Maths)|Rearrange]] the equation and [[Solve (Maths)|solve]].''' |
<math>a = \frac{12}{0.16}</math> | <math>a = \frac{12}{0.16}</math> | ||
| Line 108: | Line 108: | ||
<math>F=ma</math> | <math>F=ma</math> | ||
| − | Where | + | Where |
| − | F = The [[Resultant Force]] on the [[object]]. | + | <math>F</math> = The [[Resultant Force]] on the [[object]]. |
| − | m = The [[Inertial Mass]] of the [[object]]. | + | <math>m</math> = The [[Inertial Mass]] of the [[object]]. |
| − | a = The [[acceleration]] of the [[object]]. | + | <math>a</math> = The [[acceleration]] of the [[object]]. |
===Example Calculations=== | ===Example Calculations=== | ||
| − | ====Finding the Force given Mass and Acceleration=== | + | ====Finding the Force given Inertial Mass and Acceleration==== |
{| class="wikitable" | {| class="wikitable" | ||
| − | | style="height:20px; width: | + | | style="height:20px; width:300px; text-align:center;" |A 1.23Mg [[object]] [[accelerate]]s at a rate of 5.3x10<sup>-2</sup>m/s/s. Calculate the [[Resultant Force|resultant force]] acting on the [[object]] correct to two [[Significant Figures|significant figures]]. |
| − | | style="height:20px; width: | + | | style="height:20px; width:300px; text-align:center;" |A 1.7x10ng cell [[accelerate]]s at a rate of 320m/s/s. Calculate the [[Resultant Force|resultant force]] acting on the rocket correct to two [[Significant Figures|significant figures]]. |
|- | |- | ||
| − | | style="height:20px; width: | + | | style="height:20px; width:300px; text-align:left;" |'''1. State the known quantities''' |
m = 1.23Mg = 1230kg | m = 1.23Mg = 1230kg | ||
a = 5.3x10<sup>-2</sup>m/s/s | a = 5.3x10<sup>-2</sup>m/s/s | ||
| − | | style="height:20px; width: | + | | style="height:20px; width:300px; text-align:left;" |'''1. State the known quantities''' |
m = 1.7ng = 1.7x10<sup>-12</sup>kg | m = 1.7ng = 1.7x10<sup>-12</sup>kg | ||
| Line 133: | Line 133: | ||
a = 320m/s/s. | a = 320m/s/s. | ||
|- | |- | ||
| − | | style="height:20px; width: | + | | style="height:20px; width:300px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers into the [[equation]] and [[Solve (Maths)|solve]].''' |
<math>F=ma</math> | <math>F=ma</math> | ||
| Line 142: | Line 142: | ||
<math>F \approx 65N</math> | <math>F \approx 65N</math> | ||
| − | | style="height:20px; width: | + | | style="height:20px; width:300px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers into the [[equation]] and [[Solve (Maths)|solve]].''' |
<math>F=ma</math> | <math>F=ma</math> | ||
| Line 153: | Line 153: | ||
|} | |} | ||
| − | ====Finding the Acceleration given Mass and Force=== | + | ====Finding the Acceleration given Inertial Mass and Force==== |
{| class="wikitable" | {| class="wikitable" | ||
| − | | style="height:20px; width: | + | | style="height:20px; width:300px; text-align:center;" |A 9.2g [[object]] is subjected to a [[Resultant Force|resultant force]] of 3.7kN. Calculate the [[acceleration]] of the [[object]] correct to two [[Significant Figures|significant figures]]. |
| − | | style="height:20px; width: | + | | style="height:20px; width:300px; text-align:center;" |A 333 tonne passenger plane experiences a [[Resultant Force|resultant force]] of 1.008MN. Calculate the [[acceleration]] of the passenger plane correct to two [[Significant Figures|significant figures]]. |
|- | |- | ||
| − | | style="height:20px; width: | + | | style="height:20px; width:300px; text-align:left;" |'''1. State the known quantities''' |
m = 9.2g = 0.0092kg | m = 9.2g = 0.0092kg | ||
F = 3.7kN = 3700N | F = 3.7kN = 3700N | ||
| − | | style="height:20px; width: | + | | style="height:20px; width:300px; text-align:left;" |'''1. State the known quantities''' |
m = 333 tonne = 3.33 x 10<sup>5</sup>kg | m = 333 tonne = 3.33 x 10<sup>5</sup>kg | ||
| Line 169: | Line 169: | ||
F = 1.008MN = 1.008 x 10<sup>6</sup>N | F = 1.008MN = 1.008 x 10<sup>6</sup>N | ||
|- | |- | ||
| − | | style="height:20px; width: | + | | style="height:20px; width:300px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers and [[Evaluate (Maths)|evaluate]].''' |
<math>F=ma</math> | <math>F=ma</math> | ||
<math>3700=0.0092a</math> | <math>3700=0.0092a</math> | ||
| − | | style="height:20px; width: | + | | style="height:20px; width:300px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers and [[Evaluate (Maths)|evaluate]].''' |
<math>F=ma</math> | <math>F=ma</math> | ||
| Line 180: | Line 180: | ||
<math>1.008 \times 10^6 = (3.33 \times 10^5 )a</math> | <math>1.008 \times 10^6 = (3.33 \times 10^5 )a</math> | ||
|- | |- | ||
| − | | style="height:20px; width: | + | | style="height:20px; width:300px; text-align:left;" |'''3. [[Rearrange (Maths)|Rearrange]] the equation and [[Solve (Maths)|solve]].''' |
<math>a = \frac{3700}{0.0092}</math> | <math>a = \frac{3700}{0.0092}</math> | ||
| Line 186: | Line 186: | ||
<math>a \approx 4.0 \times 10^5m/s/s</math> | <math>a \approx 4.0 \times 10^5m/s/s</math> | ||
| − | | style="height:20px; width: | + | | style="height:20px; width:300px; text-align:left;" |'''3. [[Rearrange (Maths)|Rearrange]] the equation and [[Solve (Maths)|solve]].''' |
<math>a = \frac{1.008 \times 10^6}{3.33 \times 10^5}</math> | <math>a = \frac{1.008 \times 10^6}{3.33 \times 10^5}</math> | ||
| Line 193: | Line 193: | ||
<math>a \approx 3.0m/s/s</math> | <math>a \approx 3.0m/s/s</math> | ||
|} | |} | ||
| + | |||
| + | ====Finding the Inertial Mass given the Force and Acceleration==== | ||
| + | {| class="wikitable" | ||
| + | | style="height:20px; width:300px; text-align:center;" |An [[object]] is subjected to a [[Resultant Force|resultant force]] of 92N and [[accelerate]]s at a rate of 0.42m/s/s. Calculate the [[Inertial Mass|inertial mass]] of the [[object]] correct to two [[Significant Figures|significant figures]]. | ||
| + | | style="height:20px; width:300px; text-align:center;" |The brakes of a car provide a [[force]] of 12kN and are able to [[decelerate]] it at a rate of 8.7m/s/s. Calculate the [[Inertial Mass|intertial mass]] of the car correct to two [[Significant Figures|significant figures]]. | ||
| + | |- | ||
| + | | style="height:20px; width:300px; text-align:left;" |'''1. State the known quantities''' | ||
| + | |||
| + | a = 0.42m/s/s | ||
| + | |||
| + | F = 92N | ||
| + | | style="height:20px; width:300px; text-align:left;" |'''1. State the known quantities''' | ||
| + | |||
| + | a = 8.7m/s/s | ||
| + | |||
| + | F = 12kN = 12 x 10<sup>3</sup>N | ||
| + | |- | ||
| + | | style="height:20px; width:300px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers and [[Evaluate (Maths)|evaluate]].''' | ||
| + | |||
| + | <math>F=ma</math> | ||
| + | |||
| + | <math>92=0.42m</math> | ||
| + | | style="height:20px; width:300px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers and [[Evaluate (Maths)|evaluate]].''' | ||
| + | |||
| + | <math>F=ma</math> | ||
| + | |||
| + | <math>12 \times 10^3 = 8.7m</math> | ||
| + | |- | ||
| + | | style="height:20px; width:300px; text-align:left;" |'''3. [[Rearrange (Maths)|Rearrange]] the equation and [[Solve (Maths)|solve]].''' | ||
| + | <math>m = \frac{92}{0.42}</math> | ||
| + | |||
| + | <math>m = 219.047619kg</math> | ||
| + | |||
| + | <math>m \approx 220kg</math> | ||
| + | | style="height:20px; width:300px; text-align:left;" |'''3. [[Rearrange (Maths)|Rearrange]] the equation and [[Solve (Maths)|solve]].''' | ||
| + | <math>m = \frac{12 \times 10^3}{8.7}</math> | ||
| + | |||
| + | <math>m = 1379.31034kg</math> | ||
| + | |||
| + | <math>m \approx 1400kg</math> | ||
| + | |} | ||
| + | |||
| + | ===References=== | ||
| + | ====AQA==== | ||
| + | |||
| + | :[https://www.amazon.co.uk/gp/product/019835939X/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=019835939X&linkCode=as2&tag=nrjc-21&linkId=57e96876985fc39b1a3d8a3e3dc238b6 ''Newton’s Second Law of motion, pages 144-145, GCSE Physics; Third Edition, Oxford University Press, AQA ''] | ||
| + | :[https://www.amazon.co.uk/gp/product/1782946403/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=1782946403&linkCode=as2&tag=nrjc-21&linkId=32a0abb60dff015b15b50e9b1d7b4644 ''Newton’s Second Law, pages 164, 165, GCSE Combined Science Trilogy; Physics, CGP, AQA ''] | ||
| + | :[https://www.amazon.co.uk/gp/product/1782945970/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=1782945970&linkCode=as2&tag=nrjc-21&linkId=a120d24dcc7cc7a58192069a3aafc1d2 ''Newton’s Second Law, pages 196, 197, GCSE Physics; The Complete 9-1 Course for AQA, CGP, AQA ''] | ||
| + | :[https://www.amazon.co.uk/gp/product/1782945598/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=1782945598&linkCode=as2&tag=nrjc-21&linkId=ad276ad49df77ab4b40ab4fd0fe10102 ''Newton’s Second Law, pages 212, 214, GCSE Combined Science; The Revision Guide, CGP, AQA ''] | ||
| + | :[https://www.amazon.co.uk/gp/product/1782945598/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=1782945598&linkCode=as2&tag=nrjc-21&linkId=ad276ad49df77ab4b40ab4fd0fe10103 ''Newton’s Second Law; investigating, page 214, GCSE Combined Science; The Revision Guide, CGP, AQA ''] | ||
| + | :[https://www.amazon.co.uk/gp/product/1782946403/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=1782946403&linkCode=as2&tag=nrjc-21&linkId=32a0abb60dff015b15b50e9b1d7b4644 ''Newton’s Second Law; investigating, pages 167, 168, GCSE Combined Science Trilogy; Physics, CGP, AQA ''] | ||
| + | :[https://www.amazon.co.uk/gp/product/1782945970/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=1782945970&linkCode=as2&tag=nrjc-21&linkId=a120d24dcc7cc7a58192069a3aafc1d2 ''Newton’s Second Law; investigating, pages 199, 200, GCSE Physics; The Complete 9-1 Course for AQA, CGP, AQA ''] | ||
| + | |||
| + | ====Edexcel==== | ||
| + | |||
| + | :[https://www.amazon.co.uk/gp/product/1782945741/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=1782945741&linkCode=as2&tag=nrjc-21&linkId=30da4f2178da182547b62a7329d13b57 ''Newton’s Second Law, pages 149, 151, 154, GCSE Combined Science; The Revision Guide, CGP, Edexcel ''] | ||
| + | :[https://www.amazon.co.uk/gp/product/1782945733/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=1782945733&linkCode=as2&tag=nrjc-21&linkId=2a2dbec9db6bf5766c0458d908fa0a52 ''Newton’s Second Law, pages 16, 18, 21, GCSE Physics; The Revision Guide, CGP, Edexcel ''] | ||
| + | :[https://www.amazon.co.uk/gp/product/1292120223/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=1292120223&linkCode=as2&tag=nrjc-21&linkId=068ecf40278c32406a7f1c6e66751417 ''Newton’s Second Law, pages 18-19, GCSE Physics, Pearson Edexcel ''] | ||
| + | :[https://www.amazon.co.uk/gp/product/1292120193/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=1292120193&linkCode=as2&tag=nrjc-21&linkId=572df39392fb4200db8391d98ae6314e ''Newton’s Second Law, pages 302-303, GCSE Combined Science, Pearson Edexcel ''] | ||
| + | :[https://www.amazon.co.uk/gp/product/1782948163/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=1782948163&linkCode=as2&tag=nrjc-21&linkId=0fdbfd5dd397d6e24a9dfb250f08587f ''Newton’s Second Law, pages 35, 36, 46, GCSE Physics, CGP, Edexcel ''] | ||
| + | :[https://www.amazon.co.uk/gp/product/1782948163/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=1782948163&linkCode=as2&tag=nrjc-21&linkId=0fdbfd5dd397d6e24a9dfb250f08587f ''Newton’s Second Law; investigating, pages 39, 40, GCSE Physics, CGP, Edexcel ''] | ||
Latest revision as of 16:16, 23 November 2019
Contents
Key Stage 4 Foundation
Meaning
Newton's Second Law states that "Force = Mass x Acceleration, which means the acceleration of an object is directly proportional to the resultant force acting upon it."
About Newton's Second Law
- Newton's Second Law can be used to calculate the acceleration of an object given its mass and the resultant force acting upon it.
Equation
Force = Mass x Acceleration
\(F=ma\)
Where
\(F\) = The Resultant Force on the object.
\(m\) = The mass of the object.
\(a\) = The acceleration of the object.
Example Calculations
Finding the Force given Mass and Acceleration
| A 2.3kg object accelerates at a rate of 8.8m/s/s. Calculate the resultant force acting on the object correct to two significant figures. | A 5.5x104kg rocket accelerates at a rate of 61m/s/s. Calculate the resultant force acting on the rocket correct to two significant figures. |
| 1. State the known quantities
m = 2.3kg a = 8.8m/s/s |
1. State the known quantities
m = 5.5x104kg a = 61m/s/s |
| 2. Substitute the numbers into the equation and solve.
\(F=ma\) \(F=2.3 \times 8.8\) \(F=20.24N\) \(F \approx 20N\) |
2. Substitute the numbers into the equation and solve.
\(F=ma\) \(F=5.5 \times 10^4 \times 61\) \(F=3355000N\) \(F \approx 3.4 \times 10^6 N\) |
Finding the Acceleration given Mass and Force
| A 7kg object is subjected to a resultant force of 53N. Calculate the acceleration of the object correct to two significant figures. | A 160g snooker ball experiences a resultant force of 12N. Calculate the acceleration of the snooker ball correct to two significant figures. |
| 1. State the known quantities
m = 7kg F = 53N |
1. State the known quantities
m = 160g = 0.16kg F = 12N |
| 2. Substitute the numbers and evaluate.
\(F=ma\) \(53=7a\) |
2. Substitute the numbers and evaluate.
\(F=ma\) \(12=0.16a\) |
| 3. Rearrange the equation and solve.
\(a = \frac{53}{7}\) \(a = 7.571m/s/s\) \(a \approx 7.6m/s/s\) |
3. Rearrange the equation and solve.
\(a = \frac{12}{0.16}\) \(a = 75m/s/s\) |
Key Stage 4 Higher
Meaning
Newton's Second Law states that "Force = Mass x Acceleration, which means the acceleration of an object is directly proportional to the resultant force acting upon it."
About Newton's Second Law
- Newton's Second Law can be used to calculate the acceleration of an object given its mass and the resultant force acting upon it.
- Newton's Second Law provides a definition for inertial mass as the ratio of force to the acceleration of an object \(m= \frac{F}{a}\).
Equation
Force = (Inertial Mass) x Acceleration
\(F=ma\)
Where
\(F\) = The Resultant Force on the object.
\(m\) = The Inertial Mass of the object.
\(a\) = The acceleration of the object.
Example Calculations
Finding the Force given Inertial Mass and Acceleration
| A 1.23Mg object accelerates at a rate of 5.3x10-2m/s/s. Calculate the resultant force acting on the object correct to two significant figures. | A 1.7x10ng cell accelerates at a rate of 320m/s/s. Calculate the resultant force acting on the rocket correct to two significant figures. |
| 1. State the known quantities
m = 1.23Mg = 1230kg a = 5.3x10-2m/s/s |
1. State the known quantities
m = 1.7ng = 1.7x10-12kg a = 320m/s/s. |
| 2. Substitute the numbers into the equation and solve.
\(F=ma\) \(F=1230 \times 5.3 \times 10^{-2}\) \(F=65.19\) \(F \approx 65N\) |
2. Substitute the numbers into the equation and solve.
\(F=ma\) \(F=1.7 \times 10^{-12} \times 320\) \(F=0.000000000544N\) \(F \approx 5.4 \times 10^{-10}N\) |
Finding the Acceleration given Inertial Mass and Force
| A 9.2g object is subjected to a resultant force of 3.7kN. Calculate the acceleration of the object correct to two significant figures. | A 333 tonne passenger plane experiences a resultant force of 1.008MN. Calculate the acceleration of the passenger plane correct to two significant figures. |
| 1. State the known quantities
m = 9.2g = 0.0092kg F = 3.7kN = 3700N |
1. State the known quantities
m = 333 tonne = 3.33 x 105kg F = 1.008MN = 1.008 x 106N |
| 2. Substitute the numbers and evaluate.
\(F=ma\) \(3700=0.0092a\) |
2. Substitute the numbers and evaluate.
\(F=ma\) \(1.008 \times 10^6 = (3.33 \times 10^5 )a\) |
| 3. Rearrange the equation and solve.
\(a = \frac{3700}{0.0092}\) \(a = 402173.913m/s/s\) \(a \approx 4.0 \times 10^5m/s/s\) |
3. Rearrange the equation and solve.
\(a = \frac{1.008 \times 10^6}{3.33 \times 10^5}\) \(a = 3.027m/s/s\) \(a \approx 3.0m/s/s\) |
Finding the Inertial Mass given the Force and Acceleration
| An object is subjected to a resultant force of 92N and accelerates at a rate of 0.42m/s/s. Calculate the inertial mass of the object correct to two significant figures. | The brakes of a car provide a force of 12kN and are able to decelerate it at a rate of 8.7m/s/s. Calculate the intertial mass of the car correct to two significant figures. |
| 1. State the known quantities
a = 0.42m/s/s F = 92N |
1. State the known quantities
a = 8.7m/s/s F = 12kN = 12 x 103N |
| 2. Substitute the numbers and evaluate.
\(F=ma\) \(92=0.42m\) |
2. Substitute the numbers and evaluate.
\(F=ma\) \(12 \times 10^3 = 8.7m\) |
| 3. Rearrange the equation and solve.
\(m = \frac{92}{0.42}\) \(m = 219.047619kg\) \(m \approx 220kg\) |
3. Rearrange the equation and solve.
\(m = \frac{12 \times 10^3}{8.7}\) \(m = 1379.31034kg\) \(m \approx 1400kg\) |
References
AQA
- Newton’s Second Law of motion, pages 144-145, GCSE Physics; Third Edition, Oxford University Press, AQA
- Newton’s Second Law, pages 164, 165, GCSE Combined Science Trilogy; Physics, CGP, AQA
- Newton’s Second Law, pages 196, 197, GCSE Physics; The Complete 9-1 Course for AQA, CGP, AQA
- Newton’s Second Law, pages 212, 214, GCSE Combined Science; The Revision Guide, CGP, AQA
- Newton’s Second Law; investigating, page 214, GCSE Combined Science; The Revision Guide, CGP, AQA
- Newton’s Second Law; investigating, pages 167, 168, GCSE Combined Science Trilogy; Physics, CGP, AQA
- Newton’s Second Law; investigating, pages 199, 200, GCSE Physics; The Complete 9-1 Course for AQA, CGP, AQA
Edexcel
- Newton’s Second Law, pages 149, 151, 154, GCSE Combined Science; The Revision Guide, CGP, Edexcel
- Newton’s Second Law, pages 16, 18, 21, GCSE Physics; The Revision Guide, CGP, Edexcel
- Newton’s Second Law, pages 18-19, GCSE Physics, Pearson Edexcel
- Newton’s Second Law, pages 302-303, GCSE Combined Science, Pearson Edexcel
- Newton’s Second Law, pages 35, 36, 46, GCSE Physics, CGP, Edexcel
- Newton’s Second Law; investigating, pages 39, 40, GCSE Physics, CGP, Edexcel