Calculating Force and Stretch in a Helical Spring

[SarahFoley]

A helical spring of spring constant 270 Nm-1 hangs vertically from a fixed support. When a 1.2 kg mass is attached to the free end of the spring the length of the spring is 30cm. What is the unstretched length of the spring? What would be the length of the spring if an additional 0.8 kg was added?

A. 26 cm; 37 cm

B. 26cm; 35cm

C. 24cm; 31cm



F = kx



this is what i keep getting for the first part of the answer. which is not one of the three answers. can someone please show me how to do this properly

270 X 0.3 = 81
81N = 1.2
81N ÷ 1.2 = 67.5
67.5 = 270 X d
D = 67.5 ÷ 270 = 25cm

 

[Doc Al]

When the 1.2 kg mass is attached, what force does it exert? Use Hooke's law (F = kx) to determine how much that force stretches the spring from its unstretched length.