{"id":1877,"date":"2018-04-17T16:34:09","date_gmt":"2018-04-17T08:34:09","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=1877"},"modified":"2023-08-09T08:16:10","modified_gmt":"2023-08-09T08:16:10","slug":"hookes-law-and-elasticity-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/hookes-law-and-elasticity-problems-and-solutions.htm","title":{"rendered":"Hooke’s law and elasticity \u2013 problems and solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

Hooke’s law and elasticity \u2013 problems and solutions<\/span><\/p>\n

The change in length<\/b><\/span><\/p>\n

1. A rod has a length of L, pulled by a force of F. The amount of elongation is \u2206L. What is the magnitude of the force if the change in length is 4\u2206L.<\/span><\/p>\n

Known :<\/u><\/span><\/p>\n

Force 1 (F1<\/sub>) = F<\/span><\/p>\n

The change in length 1 (\u2206L1<\/sub>) = \u2206L<\/span><\/p>\n

The change in length 2 (\u2206L2<\/sub>) = 4 \u2206L <\/span><\/p>\n

Wanted :<\/u> Force 2 (F2<\/sub>)<\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

The equation of Hooke’s law<\/a> <\/span><\/p>\n

k = F \/ \u0394L<\/span><\/p>\n

k = constant of elasticity, F = force of F, \u0394L = the change in length <\/i><\/span><\/p>\n

k1 <\/sub>= k2<\/sub><\/span><\/p>\n

F1<\/sub> \/ \u2206L1<\/sub> = F2 <\/sub>\/ \u2206L2<\/sub><\/span><\/p>\n

F \/ \u0394L = F2 <\/sub>\/ 4\u0394L <\/span><\/p>\n

F \/ 1 = F2 <\/sub>\/ 4<\/span><\/p>\n

F = F2 <\/sub>\/ 4<\/span><\/p>\n

F2 <\/sub>= 4F<\/span><\/p>\n

2. \"Hooke'sThere springs are connected in series-parallel, as shown in figure below. Spring 1 has constant 200 N\/m, spring 2 has constant 200 N\/m and spring 3 has constant 200 N\/m. The mass of object is 100 gram and acceleration due to gravity<\/a> is 10 m\/s2<\/sup>. What is the change in length of the <\/span>equivalent spring.<\/span><\/span><\/p>\n

Known :<\/u><\/span><\/p>\n

Object’s mass<\/a> (m) = 100 gram = 0.1 kg<\/span><\/p>\n

k<\/span>1<\/sub><\/span> = k<\/span>2<\/sub><\/span> = k<\/span>3<\/sub><\/span> = 200 N\/m<\/span><\/span><\/p>\n

w = m g = (0.1 kg)(10 m\/s<\/span>2<\/sup><\/span>) = 1 kg m\/s<\/span>2<\/sup><\/span> = 1 Newton<\/span><\/span><\/p>\n

Wanted :<\/u><\/span> The change in length of the <\/span>equivalent spring.<\/span><\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

\"Hooke'sDetermine the equivalent spring constant :<\/span><\/span><\/p>\n

Spring 2 (k2<\/sub>) and spring 3 (k3<\/sub>) are connected in parallel. The equivalent spring constant :<\/span><\/span><\/p>\n

kp<\/sub> = k2<\/sub> + k3<\/sub> = 200 + 200 = 400 Nm\u22121<\/sup><\/span><\/p>\n

Spring 1 (k1<\/sub>) and spring p (kP<\/sub>) are connected in series. The equivalent spring constant :<\/span><\/p>\n

1\/ks<\/sub> = 1\/kp<\/sub> + 1\/k1<\/sub> = 1\/400 + 1\/200 = 1\/400 + 2\/400 = 3\/400<\/span><\/p>\n

ks<\/sub> = 400\/3 Nm\u22121<\/sup><\/span><\/p>\n

The equivalent spring constant is <\/span>400\/3 Nm-1<\/sup><\/span><\/p>\n

Determine the change in length of the <\/span>equivalent spring :<\/span><\/u><\/span><\/p>\n

The equation of Hooke’s law :<\/span><\/p>\n

\u2206x = F \/ k = w \/ k<\/span><\/p>\n

The change in length of the <\/span>equivalent spring :<\/span><\/span><\/p>\n

\u2206x = w \/ k<\/span><\/p>\n

\u2206x = 1 : 400\/3 = 1 x 3\/400 = 3\/400 = 0.0075 m = 0.75 cm<\/span><\/p>\n

The constant of spring<\/b><\/span><\/p>\n

3. What is the constant of spring according to the data in the table below.<\/span><\/p>\n

\"Hooke's<\/span><\/p>\n

Solution :<\/span><\/p>\n

The equation of Hooke’s law :<\/span><\/p>\n

k = F \/ \u0394x<\/span><\/p>\n

Constant of spring :<\/span><\/p>\n

k = 0.98 \/ 0.0008 = 1.96 \/ 0.0016 = 2.94 \/ 0.0024 = 3.92 \/ 0.0032 = 1.225 N\/m<\/span><\/p>\n

4. Three springs are connected in series-parallel as shown in figure below. The constant of spring k1<\/sub> = k2 <\/sub>= 3 Nm\u22121<\/sup> and k3<\/sub> = 6 Nm\u22121<\/sup>. What is the constant of the equivalent of spring. <\/span><\/p>\n

Known :\"Hooke's<\/u><\/span><\/p>\n

Constant of spring 1 (k1<\/sub>) = constant of spring 2 (k2<\/sub>) = 3 Nm\u22121 <\/sup> <\/span><\/p>\n

Constant of spring 3 (k3<\/sub>) = 6 Nm\u22121 <\/sup> <\/span><\/p>\n

Wanted :<\/u> constant of the equivalent spring (k)<\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

Spring 1 (k1<\/sub>) and spring 2 (k2<\/sub>) are connected in parallel. The Constant of the equivalent spring :<\/span><\/p>\n

kp<\/sub> = k1<\/sub> + k2<\/sub> = 3 + 3 = 6 Nm\u22121<\/sup><\/span><\/p>\n

Spring p (kP<\/sub>) and spring 3 (k3 <\/sub>) are connected in series. The constant of the equivalent spring :<\/span><\/p>\n

1\/ks<\/sub> = 1\/kp<\/sub> + 1\/k 3<\/sub> = 1\/6 + 1\/3 = 1\/6 + 2\/6 = 3\/6<\/span><\/p>\n

ks<\/sub> = 6\/2 = 3 Nm\u22121<\/sup><\/span><\/p>\n

The constant of the equivalent of spring = 3 Nm\u22121<\/sup>.<\/span><\/p>\n

5. A spring with length of L, pulled by weight of w. According to data in table below, what is the constant of the equivalent spring :<\/span><\/p>\n

\"Hooke's<\/span><\/p>\n

Solution :<\/span><\/p>\n

k = F \/ \u0394x <\/i><\/span><\/p>\n

Constant of spring :<\/span><\/p>\n

k = 10 \/ 0.02 = 20 \/ 0.04 = 30 \/ 0.06 = 40 \/ 0.08 = 500 N\/m<\/span><\/p>\n

6. According to data in table below, what is the constant of the equivalent spring :<\/span><\/p>\n

\"Hooke's<\/span><\/p>\n

Solution :<\/span><\/p>\n

k = F \/ \u0394x <\/i>= w \/ \u0394x <\/i>= m g \/ \u0394x <\/i><\/span><\/p>\n

k = constant of elasticity, w = weight, m = mass, g = acceleration due to gravity, \u0394x = the change in length<\/i><\/span><\/p>\n

Spring constant :
\n<\/span><\/p>\n

k = 2 \/ 0.05 = 4 \/ 0.1 = 6 \/ 0.15 = 8 \/ 0.20 = 10 \/ 0.25 = 40 N\/m<\/span><\/p>\n

7. If k1<\/sub> = 4k, what is the constant of the equivalent spring.<\/span><\/p>\n

Solution :\"Hooke's<\/span><\/p>\n

Two springs are connected in parallel. The constant of the equivalent spring :<\/span><\/p>\n

kp<\/sub> = k + k = 2k<\/span><\/p>\n

Two springs are connected in series. The constant of the equivalent spring<\/span><\/p>\n

1\/ks<\/sub> = 1\/kp<\/sub> + 1\/k1<\/sub> = 1 \/ 2k + 1 \/ 4k = 2 \/ 4k + 1 \/ 4k = 3 \/ 4k<\/span><\/p>\n

ks<\/sub> = 4k\/3 <\/span><\/p>\n

8. According to data in table below, what is the constant of the equivalent spring :<\/span><\/p>\n

\"Hooke's<\/span><\/p>\n

Solution :<\/span><\/p>\n

The equation of Hooke’s law :<\/span><\/p>\n

k = F \/ \u0394<\/i>L <\/i><\/span><\/p>\n

Constant of spring :<\/span><\/p>\n

k = 2 \/ 0.0050 = 3 \/ 0.0075 = 4 \/ 0.01 = 400 Nm-1<\/sup><\/span><\/p>\n

9. The smallest constant is…
\n<\/span><\/p>\n

\"Hooke's<\/span><\/p>\n

Solution<\/u><\/span><\/p>\n

The equation of Hooke’s law :<\/span><\/p>\n

k = F \/ \u0394x <\/span><\/p>\n

k = constant of elasticity, F = force, \u0394x = the change in length <\/i><\/span><\/p>\n

Constant of elasticity :<\/span><\/p>\n

kA<\/sub> = F \/ \u0394x = 1 \/ 0.05 = 20 N\/m<\/span><\/p>\n

kB<\/sub> = F \/ \u0394x = 2 \/ 0.025 = 80 N\/m<\/span><\/p>\n

kC<\/sub> = F \/ \u0394x = 1 \/ 0.025 = 40 N\/m <\/span><\/p>\n

kD<\/sub> = F \/ \u0394x = 2 \/ 0.05 = 40 N\/m<\/span><\/p>\n

kE<\/sub> = F \/ \u0394x = 2 \/ 0.25 = 8 N\/m <\/span><\/p>\n

10. What is the largest constant according to data in table below.<\/span><\/p>\n

\"Hooke's<\/span><\/p>\n

Solution :<\/span><\/p>\n

The equation of the Hooke’s law :<\/span><\/p>\n

k = F \/ \u0394x <\/span><\/p>\n

kA<\/sub> = 7 \/ 0.035 = 200 Nm-1<\/sup><\/span><\/p>\n

kB<\/sub> = 8 \/ 0.025 = 320 Nm-1<\/sup><\/span><\/p>\n

kC<\/sub> = 6 \/ 0.020 = 300 Nm-1<\/sup><\/span><\/p>\n

kD<\/sub> = 9 \/ 0.045 = 200 Nm-1<\/sup><\/span><\/p>\n

kE<\/sub> = 10 \/ 0.033 = 303 Nm-1<\/sup><\/span><\/p>\n

The largest constant is 320 Nm-1<\/sup>.<\/span><\/p>\n

11. The graph below show connection between the change in force (\u0394F) and the increase in length (\u0394x). What is the graph showing the smallest constant of elasticit<\/span>y.<\/span><\/span><\/p>\n

\"Hooke's<\/span><\/p>\n

Solution<\/span><\/p>\n

The equation of Hooke’s law :<\/u><\/span><\/p>\n

k = F \/ \u0394x <\/span><\/p>\n

\u0394x = the change in length, F = force, k = constant of elasticity<\/i><\/span><\/p>\n

Constant of elasticity :<\/u><\/span><\/p>\n

kA<\/sub> = F \/ \u0394x = 1 \/ 8 = 0.125 <\/span><\/p>\n

kB<\/sub> = F \/ \u0394x = 8 \/ 3 = 2.7<\/span><\/p>\n

kC<\/sub> = F \/ \u0394x = 6 \/ 6 = 1 <\/span><\/p>\n

kD<\/sub> = F \/ \u0394x = 3 \/ 5 = 0.6<\/span><\/p>\n

kE<\/sub> = F \/ \u0394x = 2 \/ 4 = 0.5<\/span><\/p>\n

12. Which grap<\/span>h <\/span>has the largest elastic constants?<\/span><\/span><\/p>\n

\"Hooke's<\/span><\/p>\n

Solution :<\/span><\/p>\n

Constant of elasticity<\/a> :<\/u><\/span><\/p>\n

kA<\/sub> = F \/ \u0394x = 50 \/ 10 = 5<\/span><\/p>\n

kB<\/sub> = F \/ \u0394x = 50 \/ 0.1 = 500<\/span><\/p>\n

kC<\/sub> = F \/ \u0394x = 5 \/ 0.1 = 50 <\/span><\/p>\n

kD<\/sub> = F \/ \u0394x = 500 \/ 0.1 = 5000<\/span><\/p>\n

kE<\/sub> = F \/ \u0394x = 500 \/ 10 = 50<\/span><\/p>\n

The potential energy of spring :<\/b><\/span><\/p>\n

13.The graph below shows the relationship between force and <\/span>the change in <\/span>spring length. <\/span>What is the potential energy of spring, according to the graph below.<\/span><\/span><\/p>\n

Known :\"Hooke's<\/u><\/span><\/p>\n

F = 40 N<\/span><\/p>\n

x = 0.08 meters<\/span><\/p>\n

Wanted <\/u>:<\/u> The potential energy of spring<\/a><\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

Constant of spring :<\/span><\/p>\n

k = F \/ \u0394x = 40 \/ 0.08 = 500 N\/m<\/span><\/p>\n

The potential energy of spring :<\/span><\/p>\n

PE = 1\/2 k x2 <\/sup>= 1\/2 (500)(0.08) = (250)(0.08) = 20 Joule<\/span><\/p>\n

14. A 2-kg block is attached at spring. If the increase in length of spring is 5 cm and acceleration due to gravity is 10 m\/s2<\/sup>, what is the potential energy of spring.<\/span><\/p>\n

Known :\"Hooke's<\/u><\/span><\/p>\n

The increase in length (\u0394x) = 5 cm = 0.05 meter<\/span><\/p>\n

Acceleration due to gravity (g) = 10 m\/s2<\/sup><\/span><\/p>\n

Block’s mass (m) = 2 kg<\/span><\/p>\n

Block’s weight (w) = m g = (2)(10) = 20 Newton<\/span><\/p>\n

Wanted :<\/u> the potential energy of spring<\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

The constant of elasticity :<\/span><\/p>\n

k = w \/ \u0394x = 20 \/ 0.05 = 400 N\/m<\/span><\/p>\n

The potential energy of spring :<\/span><\/p>\n

PE = \u00bd k \u0394x2 <\/sup>= \u00bd (400)(0.05)2<\/sup> = (200)(0.0025) <\/span><\/p>\n

PE = 0.5 Joule <\/span><\/p>\n

15. The change in length of spring is 5 cm when pulled by 20-N force. What is the potential energy of spring when the change in length of the spring is 10 cm.<\/span><\/p>\n

Known :<\/u><\/span><\/p>\n

The change in length (\u0394x) = 5 cm = 0.05 meters<\/span><\/p>\n

Force (F) = 20 Newton<\/span><\/p>\n

Wanted <\/u>:<\/u> The potential energy of spring<\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

The constant of spring :<\/span><\/p>\n

k = F \/ \u0394x = 20 \/ 0.05 = 400 N\/m<\/span><\/p>\n

The potential energy of spring when \u0394x = 10 cm = 0.1 m :<\/span><\/p>\n

PE = \u00bd k \u0394x2 <\/sup>= \u00bd (400)(0.1)2<\/sup> = (200)(0.01) <\/span><\/p>\n

PE = 2 Joule<\/span><\/p>\n

Object’s weight <\/b><\/span><\/p>\n

16. Four springs where the constant of each spring is 800 N\/m, connected in series-parallel, as shown in figure. A block is attached at spring. The change in length of all springs is 5 cm. What is the weight of block.<\/span><\/p>\n

Known :\"Hooke's<\/u><\/span><\/p>\n

k1<\/sub> = k2<\/sub> = k3<\/sub> = k4 <\/sub>= 800 N m\u22121<\/sup><\/span><\/p>\n

\u0394x = 5 cm = 0.05 m<\/span><\/p>\n

Wanted : <\/u> block’s weight (w)<\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

Determine the constant of the equivalent spring<\/u><\/span><\/p>\n

\"Hooke'sSpring 1 (k1<\/sub>), spring 2 (k2<\/sub>) and spring 3 (k3<\/sub>) are connected in parallel. The constant of the equivalent spring :<\/span><\/p>\n

kp<\/sub> = k1<\/sub> + k2<\/sub> + k3<\/sub> = 800 + 800 + 800 = 2400 Nm\u22121<\/sup><\/span><\/p>\n

Spring p (kP<\/sub>) and spring 4 (k4<\/sub>) are connected in series. The constant of the equivalent spring :<\/span><\/p>\n

1\/ks<\/sub> = 1\/kp<\/sub> + 1\/k4<\/sub> = 1\/2400 + 1\/800 = 1\/2400 + 3\/2400 = 4\/2400<\/span><\/p>\n

ks<\/sub> = 2400\/4 = 600 Nm\u22121<\/sup><\/span><\/p>\n

The constant of the equivalent spring is 600 Nm-1<\/sup><\/span><\/p>\n

Determine the weight of object :<\/u><\/span><\/p>\n

The equation of Hooke’s law :<\/span><\/p>\n

F = k \u0394x tau w = k \u0394x <\/span><\/p>\n

Weight of object :<\/span><\/p>\n

w = (600 Nm-1<\/sup>)(0.05 m) = 30 Newton<\/span><\/p>\n

17. Four springs are connected in series-parallel. Constant of each spring is 1600 N\/m. A block is attached at the end of spring, as shown in figure. The increase in length of all spring is 5 cm. What is the weight of block.<\/span><\/p>\n

Known :\"Hooke's<\/u><\/span><\/p>\n

k1<\/sub> = k2<\/sub> = k3<\/sub> = k4 <\/sub>= 1600 N m\u22121<\/sup><\/span><\/p>\n

\u0394x = 5 cm = 0.05 m<\/span><\/p>\n

Wanted :<\/u> weight of block<\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

Determine the constant of the equivalent spring<\/u><\/span><\/p>\n

\"Hooke'sSpring 1 (k1<\/sub>), spring 2 (k2<\/sub>) and spring 3 (k3<\/sub>) are connected in parallel. The constant of the equivalent spring :<\/span><\/p>\n

kP<\/sub> = k1<\/sub> + k2<\/sub> + k3<\/sub> = 1600 + 1600 + 1600 = 4800 Nm\u22121<\/sup><\/span><\/p>\n

Spring p (kP<\/sub>) and spring 4 (k4<\/sub>) are connected in series. The constant of the equivalent spring :<\/span><\/p>\n

1\/ks<\/sub> = 1\/kp<\/sub> + 1\/k4<\/sub> = 1\/4800 + 1\/1600 = 1\/4800 + 3\/4800 = 4\/4800<\/span><\/p>\n

ks<\/sub> = 4800\/4 = 1200 Nm\u22121<\/sup><\/span><\/p>\n

The constant of the equivalent spring is 1200 Nm-1<\/sup><\/span><\/p>\n

Determine the weight of object :<\/u><\/span><\/p>\n

The equation of Hooke’s law :<\/span><\/p>\n

F = k \u0394x or w = k \u0394x<\/span><\/p>\n

Object’s weight :<\/span><\/p>\n

w = (1200 Nm-1<\/sup>)(0.05 m) = 60 Newton<\/span><\/p>\n

    \n
  1. What is Hooke’s Law?<\/strong><\/span>\n