= 15 N (ANS) T = 4 × 2 Note that this question was done very poorly on the T = 8 N (ANS) exam. Question 28 (2016 Q3a, 2m, 60%) Question 25 (2014 Q1b, 2m, 50%) Use F = k Δx Use T = With 1 mass, F = 0.050 × 9.8 ∑ma = 2 × 10 × 103 × 0.2 F = 0.49 N. = 4 × 103 N (ANS) 0.49 = k × 0.25 k = 1.96 k = 2 N m-1 (ANS) Question 26 (2014 Q2a, 2m, 60%) Use the equation F = k × ∆x Question 29 (2016 Q4d, 3m, 57%) F = mg, so 0.05 × 10 = k × 0.1 Use the work done is the change in KE. 0.5 / 0.1 = k F × d = ΔKE k = 5 N m-1 KEi = mv2 = × 4 × 22 0.1 × 10 = k × 0.2 = 8 J 0.1 / 0.2 = k F × d = 8 k = 5 N m-1 2.0 × d = 8 d = 4 m (ANS) 0.15 × 10 = k × 0.3 0.15 / 0.3 = k k = 5 N m-1 Question 30 (~2016 Q1) k = 5 N m-1 (ANS) For the wagon to accelerate at 0.10 m s-1, use F = ma, F is the tension in the coupling. Question 27a (2015 Q2a, 2m, 52%) F = 10 000 × 0.10 F = 1000 N (ANS) For M2, use = ma ΣF m2g T = m2a – 1 × 10 T = 1 × a – For M1, use = ma ΣF T = m1a T = 4 × a Combine the two equations to get 10 4a = a – 10 = 5a a = 2 m s-2 (ANS) Question 27b (2015 Q2b, 2m, 54%) Consider M1, T = 4 × a
= 15 N (ANS) Note that this question was done very poorly on the exam. Question 25 (2014 Q1b, 2m, 50%) Use T = ∑ma = 2 × 10 × 103 × 0.2 = 4 × 103 N (ANS) Question 26 (2014 Q2a, 2m, 60%) Use the equation F = k × ∆x F = mg, so 0.05 × 10 = k × 0.1 0.5 / 0.1 = k k = 5 N m-1 0.1 × 10 = k × 0.2 0.1 / 0.2 = k k = 5 N m-1 0.15 × 10 = k × 0.3 0.15 / 0.3 = k k = 5 N m-1 k = 5 N m-1 (ANS) Question 27a (2015 Q2a, 2m, 52%) For M2, use ΣF = ma m2g – T = m2a 1 × 10 – T = 1 × a For M1, use ΣF = ma T = m1a T = 4 × a Combine the two equations to get 10 – 4a = a 10 = 5a a = 2 m s-2 (ANS) Question 27b (2015 Q2b, 2m, 54%) Consider M1, T = 4 × a T = 4 × 2 T = 8 N (ANS) Question 28 (2016 Q3a, 2m, 60%) Use F = k Δx With 1 mass, F = 0.050 × 9.8 F = 0.49 N. 0.49 = k × 0.25 k = 1.96 k = 2 N m-1 (ANS) Question 29 (2016 Q4d, 3m, 57%) Use the work done is the change in KE. F × d = ΔKE KEi = mv2 = × 4 × 22 = 8 J F × d = 8 2.0 × d = 8 d = 4 m (ANS) Question 30 (~2016 Q1) For the wagon to accelerate at 0.10 m s-1, use F = ma, F is the tension in the coupling. F = 10 000 × 0.10 F = 1000 N (ANS)
2.-Checkpoints-Forces-solutions.docx