Thats what I got so hopefully others are wrong

Edit: Actually I think I got 15, 35 oops

Q19. Most part of the question 19 is ok. (total 18 marks)

Plotting the graph is straight forward (just to time mass in kg x 10 to convert to force, The compression in mm in x axis.

When you plot the graph you'll get two linear sections. Gradient of first linear graph is the K of spring A (which is 150 N/m). The gradient of the second liner section is the sum of K values of A and B which was 500 N/m. Hence the K of the spring B = 350 N/m. The scales need to be selected to spread your data more than 50% of the area. Uncertainty in compression was +/- 2.0mm . If you have selected scales appropriately, the error bars are like half a box left and right.

(c) (i) Area under the graph of spring A = 1/2*150*0.08^2 = 0.48J

C(ii) Spring Pot energy stored in the spring when the system is compressed by 80 mm. Need to calculate the area under both straight lines= 1/2*0.06*9 + (1/2(9+19)*0.02 =0.55 J

(iii) Work done only by spring B to compress to 80 mm = 1/2 * 350*0.02^2 (because Spring B only compresses from 60 mm to 80 mm). = 0.07 J

(b) How these type spring system can be used in small bumps and severe bumps.

Low K spring for small bumps and combination of Both springs for sever bumps to absorbs the shock.