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junior college 1 | H2 Maths
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#H1Physics
Hello! Can someone kindly help me out with this whole question?
for (a), i can’t figure out the area.
for (b), i can only find the constant k in F=kx...
so i can’t carry on:( Thank u so much
Date Posted: 4 years ago
Views: 295
Good evening Sonia!
Since F = kx, where F is the force needed to pull the spring and x is the displacement from its unstretched position (at 6 cm), the area of the graph representing energy stored is the triangle formed by (0, 6), (5, 6) and (5, 10).
When the mass is suspended, it automatically goes to the 9.2 cm mark of the length.
Pulling a further 0.80 cm means that its new total length is now 10.0 cm.
Change in EPE
= The shaded area from x = 4.0 to x = 5.0
= Area of trapezium (no need integrals!)
= 1/2 * 1.0 * [(3.2 + 4.0)]/100
= 0.036 J
(divide by 100 means change cm to m)
Since F = kx, where F is the force needed to pull the spring and x is the displacement from its unstretched position (at 6 cm), the area of the graph representing energy stored is the triangle formed by (0, 6), (5, 6) and (5, 10).
When the mass is suspended, it automatically goes to the 9.2 cm mark of the length.
Pulling a further 0.80 cm means that its new total length is now 10.0 cm.
Change in EPE
= The shaded area from x = 4.0 to x = 5.0
= Area of trapezium (no need integrals!)
= 1/2 * 1.0 * [(3.2 + 4.0)]/100
= 0.036 J
(divide by 100 means change cm to m)
See 1 Answer
done
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A very good evening to you Sonia! Here are my workings for this question.
Date Posted:
4 years ago
Thank u so much!!! Have a great saturday :)
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