Question 5 of 6: Question 5 (paper Question V) — Spring-Restrained Cart on an Incline (Part B · Dynamics, 20 marks)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Examination December 2015 — 04-BS-3, Statics and Dynamics, 3 hours, closed book (one 8½″×11″ self-prepared note sheet permitted; Casio or Sharp approved calculator only). Candidates complete 2 of 3 questions from Part A (Statics) and 2 of 3 from Part B (Dynamics); all 6 are solved below.
Check — Figure 1 is drawn on a grid of 1 division = 1 m; node positions are measured against it, and each figure's geometry is stated at first use. All six printed questions (I–VI) are answered below.
Question 5 (paper Question V) — Spring-Restrained Cart on an Incline (Part B · Dynamics, 20 marks)
Given. m = 10 kg; θ = 20°; frictionless incline; one spring up-slope of the cart and two springs down-slope, all k = 125 N/m, all unstretched at x = 0; cart released from rest at x = 0; x measured positive down the slope.
Given data
Quantity
Value
m
10 kg
θ
20°
k (each spring)
125 N/m
Number of springs engaged for any x≠0
3 (1 stretches, 2 compress)
Initial condition
x=0, v=0
Find. (a) speed at x = 50 mm; (b) xmax; (c) final (equilibrium) position.
Figure 5 — cart on a 20° frictionless incline restrained by one spring above and two below.
Approach. Because all three springs are unstretched together at x = 0 and both the up-slope spring (which stretches) and the two down-slope springs (which compress) resist any positive displacement x, they act as three parallel springs with a single effective stiffness $k_{eff}=k+2k=3k$; the work–energy theorem then gives speed directly at any x, and the turning points / equilibrium follow from the same energy balance and from Newton's second law.
Effective stiffness and driving force.
$$k_{eff}=k+2k=3(125)=375\ \text{N/m}, \qquad mg\sin\theta=10(9.81)\sin20^\circ=33.55\ \text{N}$$
Speed at x = 0.05 m (work–energy theorem). Gravity does positive work $mg\sin\theta\,x$ down-slope while the springs store $\tfrac12 k_{eff}x^2$:
$$\tfrac12 mv^2 = mg\sin\theta\,x-\tfrac12k_{eff}x^2$$
$$v=\sqrt{2g\sin\theta\,x-\dfrac{k_{eff}}{m}x^2}=\sqrt{2(9.81)(0.34202)(0.05)-37.5(0.05)^2}=\boxed{0.492\ \text{m/s}}$$
Maximum displacement (v = 0 again). Setting the same energy expression to zero at the far turning point (x≠0):
$$2mg\sin\theta\,x_{max}=k_{eff}x_{max}^2 \;\Rightarrow\; x_{max}=\dfrac{2mg\sin\theta}{k_{eff}}=\dfrac{2(33.55)}{375}=\boxed{0.1789\ \text{m} = 178.9\ \text{mm}}$$
Final (equilibrium) position. The cart oscillates in simple harmonic motion about the point where the net force is zero, $mg\sin\theta=k_{eff}x_{eq}$; the motion described is undamped SHM between x = 0 and x = xmax, and its centre — the position about which it settles once any (unstated) damping removes the oscillation energy — is exactly $x_{max}/2$:
$$x_{eq}=\dfrac{mg\sin\theta}{k_{eff}}=\dfrac{33.55}{375}=\boxed{0.0895\ \text{m} = 89.5\ \text{mm}}$$
