Questions
- The following results were recorded in an experiment where different masses were hung on the end of a long spring whose other end was firmly fixed. The length of the spring and the mass hanging from it were recorded as below. Original length of spring was 40cm.
Length of spring (cm) 44 48 52 56 60 65 70 74 Mass attached (kg) 0.15 0.30 0.45 0.60 0.75 0.90 1.05 1.20 - Complete the table for load and corresponding extensions
- Plot a graph of extension of the spring against load on the spring on the grid provided
- Determine the spring constant using the linear section of the graph
- Give an explanation why the slope of the graph changes when a mass greater than 0.75kg is attached to the spring
- From the list of quantities below, select quantities that are vector quantities:- speed, density, force, acceleration and current
- Sketch a graph of length of a helical spring against compressing force until the coils of the spring are in contact
- The three springs shown in the figure below are identical and have negligible weight. The extension produced on the system of springs is 20cm
Determine the constant of each spring - The graphs in the figure below represents the relations between extension e and mass, m added on two springs x and y
Given that the two springs are made from the same material, give a reason why the graphs are different - A single light spring extends by 3.6cm when supporting a load of 2.5kg. What is the total extension in the arrangement shown below. (Assume the springs are identical)
- Three identical springs with proportionality constant of 50 N/m each are connected as shown below and support a load of 60N
Calculate;- The extension in one spring
- The extensive proportionality constant of the springs
- When a load of 20N is hung from a spring, the spring has a length of 15 cm. The same spring has a length of 17 cm when supporting a load of 25N. Determine the spring length when supporting no load.
The figure below shows a U-tube manometer. Use it to answer question 5 and 6. Density of water = 100 kgm-3. - The diagram below shows three identical springs which obey Hooke’s law.
Determine the length X
Answers
-
Mass attached (kg) 0.15 0.30 0.45 0.60 0.75 0.90 1.05 1.20 Force (load) on the spring(N) 1.5 3.0 4.5 6.0 7.5 9.0 10.5 12.0 Extension of spring (m) 0.04 0.08 0.12 0.16 0.20 0.25 0.30 0.34 -
- slope = 0.16 – 0.05 = 0.11 = 0.02683
6.0 – 1.9 4.1
Spring constant = 1/0.02683 = 37.27Nm-1 - – Force
- Acceleration
-
- F1 = Ke1 = 40 = Ke1
e1 = 40/K
F2 = Ke2 = 20 = Ke2
K K
e2 = 20
K
but e1 + e2 = 20
40 + 20 = 20cm
K K
60 = 20k
K = 3N/cm - Diameter of coils/ Thickness/ No. of turns per unit length / length of spring are different ✓1
- Upper springs, e = 3.6/3
Middle springs, e = 3.6/2 = 18cm
Lower springs, e = 3.6/1 = 3.6cm
Total extension = 1.2 + 1.8 + 3.6
= 6.6cm -
- Load on each spring = 60/3
= 20N ✓
Extension (e) in one spring = F/K ✓for one spring
= 20/50
= 0.4m ✓ - The effective constant (K)
= K1 + K2 + K3 ✓
= 3(50)
= 150 N/m ✓
- Load on each spring = 60/3
- A load of (25 – 20)N causes extension of (17 – 15) cm.
i.e. 5N causes extension of 2 cm ✓1
20N = ?
20/5 N x 2 cm = 8 cm ✓1
When no mass is hung.
Length of the spring = 15cm – 8cm
= 7 cm ✓1
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