2.5 Newton’s Second Law — Worked Solutions
Work through Topic 2.5
These are the full worked solutions to the lesson questions. Try the questions first, then check each step and the reason for it.
Question 1
A cart moves right but its net force points left. What must be true?
- Its acceleration points left
- Its velocity instantly points left
- Its acceleration is zero until it stops
Step 1. Choose right positive. A leftward net force has F_net,x < 0, so a_x = F_net,x/m < 0.
Why this step? Positive mass cannot change the net force’s sign when dividing in Newton’s second law.
Step 2. The cart may still have v_x > 0 while its velocity decreases.
Why this step? Acceleration is a rate of change, so an opposing force initially slows the cart instead of making velocity reverse instantaneously.
Step 3. The guaranteed statement is that acceleration points left.
Why this step? Whether or when the cart reverses additionally depends on the force duration and its initial rightward velocity.
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Question 2
Net force and system mass both double. How does acceleration change?
- It doubles
- It is unchanged
- It halves
Step 1. Write the original acceleration as a = F_net/M.
Why this step? The comparison concerns the resultant external force and total system mass, not one isolated force or constituent mass.
Step 2. After both double, a_new = (2F_net)/(2M) = F_net/M = a.
Why this step? The same factor multiplies numerator and denominator, so the ratio determining acceleration remains unchanged.
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Question 3
For two connected carts treated as one system, what happens to the string interaction in the force sum?
- It counts twice in the pulling direction
- Its internal pair cancels
- It is always zero on each cart
Step 1. Include both carts and their connecting ideal string in the whole-system boundary.
Why this step? This choice places the two ends of the connecting interaction inside the system rather than leaving a tension as an outside pull.
Step 2. The tension forces on the carts are equal and opposite, so their internal pair sums to zero.
Why this step? Newton’s third-law pairing cancels internal interactions in the system’s translational force sum.
Step 3. Each cart separately can still have nonzero tension in its own equation.
Why this step? Cancellation is a property of adding both recipients’ equations; it does not mean either cart is unforced.
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Question 4
A 4 kg cart has a 15 N rightward pull and 3 N leftward friction. Find acceleration and the velocity change after 2 s of constant force.
Step 1. Take right positive and add the horizontal forces: F_net = 15 − 3 = +12 N.
Why this step? Friction points left, so its component opposes rather than supplements the pull.
Step 2. Use a = F_net/m = 12/4 = +3 m/s².
Why this step? The 4 kg cart is the selected system, and the constant resultant gives constant acceleration over the stated interval.
Step 3. Find Δv = aΔt = 3·2 = +6 m/s.
Why this step? Two seconds of the same acceleration add a fixed positive velocity change; this does not require the cart to start from rest.
Step 4. Final velocity would be v_f = v_i + 6 m/s.
Why this step? Initial velocity was not provided, so reporting +6 m/s as a final velocity would assume an unstated starting condition.
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Question 5
For the two-cart example, derive a = F/(m₁ + m₂) and T = m₂F/(m₁ + m₂) on a frictionless track. Explain which system gives each equation.
Step 1. For both carts, the external horizontal force is F and total mass is m₁ + m₂, so F = (m₁ + m₂)a.
Why this step? The track is frictionless, and the connecting tension is internal to the combined system; the taut ideal string gives a common acceleration.
Step 2. Solve a = F/(m₁ + m₂).
Why this step? The external pull must accelerate both masses, so using just the front mass would overestimate the shared acceleration.
Step 3. For the rear cart alone, T = m₂a.
Why this step? Only the string supplies its horizontal force in this frictionless model, so its own mass determines the required tension.
Step 4. Substitute the shared acceleration: T = m₂F/(m₁ + m₂).
Why this step? This connects the whole-system result to the rear-cart force without reclassifying tension as an external force on both carts.
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Question 6
A best-fit graph of a versus F_net has slope 0.25 kg⁻¹. Infer the mass. Explain how an unaccounted constant friction force would affect the graph’s intercept.
Step 1. For correctly calculated net force, a = (1/M)F_net, so the fitted slope is 1/M.
Why this step? The graph places acceleration vertically and net force horizontally; its slope has inverse-mass units.
Step 2. Invert the measured slope: M = 1/(0.25 kg⁻¹) = 4 kg.
Why this step? The reciprocal converts inverse kilograms to kilograms; multiplying by the slope would return the wrong dimensions.
Step 3. If the horizontal value actually omits constant opposing friction, it is F_pull, and a = F_pull/M − f/M.
Why this step? The true net force is F_pull − f for motion in the positive direction; the omitted force appears as a constant negative offset.
Step 4. That mislabeled applied-force graph has vertical intercept −f/M; a true F_net graph retains zero intercept in this model.
Why this step? An intercept shift diagnoses unaccounted force or calibration rather than a different mass slope, provided friction stays constant in the fitted interval.