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203-AS1-MS Mechanics tutoring

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What 203-AS1-MS covers

Marianopolis’s Arts & Sciences course 203-AS1-MS covers classical mechanics: how objects move and rotate, and applications of these ideas. The question below focuses on forces and motion along a straight line.

Marianopolis January 2026 course catalogue

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A cart of mass m=2.0 kgm=2.0\,\mathrm{kg} starts from rest on a level track. A constant horizontal pull of 6.0 N6.0\,\mathrm{N} acts to the right; a constant resistive force of 2.0 N2.0\,\mathrm{N} acts to the left. Treat the cart as a particle and neglect any other horizontal forces. Find (a) its horizontal acceleration and (b) its horizontal velocity after 3.0 s3.0\,\mathrm{s}, stating the direction in each case. Then explain why using 6.0 N6.0\,\mathrm{N} alone in Newton’s second law would be wrong.

Show the reasoning

  1. Choose a direction and define the quantities

    Take right as positive. Let mm be the cart’s mass, FnetF_{\mathrm{net}} its net horizontal force, aa its horizontal acceleration, v0v_0 its initial horizontal velocity, tt the elapsed time and vv its horizontal velocity at that time. Here m=2.0 kgm=2.0\,\mathrm{kg}, v0=0 m/sv_0=0\,\mathrm{m/s} and t=3.0 st=3.0\,\mathrm{s}.

  2. Add the forces before dividing by the mass

    The horizontal forces point in opposite directions, so Fnet=+6.0 N−2.0 N=+4.0 N.\begin{aligned}F_{\mathrm{net}} &= +6.0\,\mathrm{N}-2.0\,\mathrm{N}\\&=+4.0\,\mathrm{N}.\end{aligned} Newton’s second law gives Fnet=maF_{\mathrm{net}}=ma, hence a=Fnetm=4.0 N2.0 kg=+2.0 m/s2.\begin{aligned}a &= \frac{F_{\mathrm{net}}}{m}\\&=\frac{4.0\,\mathrm{N}}{2.0\,\mathrm{kg}}\\&=+2.0\,\mathrm{m/s^2}.\end{aligned} The acceleration points right.

  3. Use the constant acceleration to find velocity

    The forces and mass are constant, so the acceleration is constant. Use v=v0+at=0 m/s+(2.0 m/s2)(3.0 s)=+6.0 m/s.\begin{aligned}v &= v_0+at\\&=0\,\mathrm{m/s}\\&\quad{}+(2.0\,\mathrm{m/s^2})(3.0\,\mathrm{s})\\&=+6.0\,\mathrm{m/s}.\end{aligned} After 3.0 s3.0\,\mathrm{s}, the cart moves right at 6.0 m/s6.0\,\mathrm{m/s}.

  4. Check the model and units

    Using 6.0 N6.0\,\mathrm{N} alone would leave out the resisting force and overestimate the acceleration. The weight and track’s normal force balance vertically in this model. Since N/kg=m/s2\mathrm{N/kg}=\mathrm{m/s^2}, the acceleration has the right units; multiplying it by seconds gives m/s\mathrm{m/s}.

Need help with the force equation? Start with free-body diagrams. For the velocity equation, review motion graphs and kinematics.

How a tutor can work through your problem

A tutor can help you choose the object to study, draw its forces and choose axes before writing an equation. If you have started a solution, you can review it together and check signs and units, then solve a similar problem yourself. For a torque question, the tutor can help you choose a pivot and find the perpendicular lever arm.

Arts & Sciences mechanics: test the model before the formula

Split a stop into two stages, test whether static friction can hold a crate, and balance a loaded beam. State the assumption behind each equation.

Before you start: Allow 20 minutes for the three questions, then 10 minutes to check the assumptions and retry one setup.

1. A stop has two stages

In a simplified model, a car moves at 12 m/s12\,\mathrm{m/s} during a 0.75 s0.75\,\mathrm s reaction interval. It then brakes with constant acceleration −3.0 m/s2-3.0\,\mathrm{m/s^2} until it stops. Find the total distance travelled from the start of the reaction interval.

Show the explained solution
  1. During reaction, acceleration is zero and the distance is dr=12(0.75)=9.0 md_r=12(0.75)=9.0\,\mathrm m. Braking has not started.
  2. During braking, use vf2=vi2+2adbv_f^2=v_i^2+2ad_b: 0=122+2(−3.0)db0=12^2+2(-3.0)d_b, giving db=24 md_b=24\,\mathrm m. This constant-acceleration equation applies to the braking stage only.
  3. The total is dr+db=33 md_r+d_b=33\,\mathrm m. Using the braking acceleration throughout the reaction interval would mix two different motion models.

Review motion in separate time intervals

2. Static friction adjusts

A 4.0 kg4.0\,\mathrm{kg} crate rests on a level floor. A horizontal push of 12 N12\,\mathrm N acts on it. The static-friction coefficient is μs=0.40\mu_s=0.40; use g=9.8 m/s2g=9.8\,\mathrm{m/s^2}. Find the actual friction force and decide whether the crate starts sliding.

Show the explained solution
  1. Vertically the crate does not accelerate, and the push has no vertical component. Therefore N=mg=39.2 NN=mg=39.2\,\mathrm N.
  2. The largest available static friction is fs,max⁡=μsN=15.68 Nf_{s,\max}=\mu_sN=15.68\,\mathrm N. Balancing the applied push needs only 12 N12\,\mathrm N, which lies below that limit.
  3. The actual friction is 12 N12\,\mathrm N, opposite the push; the crate stays at rest. The value 15.68 N15.68\,\mathrm N is a maximum, not the automatic friction force. With no sliding, a kinetic-friction law is not appropriate.

Review force diagrams and friction

3. Choose a pivot, then check forces

A uniform horizontal beam is 2.0 m2.0\,\mathrm m long and weighs 60 N60\,\mathrm N. Vertical supports act at both ends. A 40 N40\,\mathrm N load acts downward 1.5 m1.5\,\mathrm m from the left end. Find both upward support forces when the beam is at rest.

Show the explained solution
  1. The beam’s own weight acts at its centre, 1.0 m1.0\,\mathrm m from the left. Taking moments about the left end removes the unknown left support from the torque equation.
  2. Let RRR_R be the right support force. Counterclockwise positive gives 2.0RR−60(1.0)−40(1.5)=0.2.0R_R-60(1.0)-40(1.5)=0. Hence RR=60 NR_R=60\,\mathrm N. All lever arms are perpendicular to the vertical forces.
  3. Vertical force balance requires RL+RR=60+40R_L+R_R=60+40, so RL=40 NR_L=40\,\mathrm N. The supports sum to 100 N, and their different values balance the off-centre load. Equal supports would balance vertical force but leave a nonzero net torque.

Practise torque and static equilibrium

Use your answers to plan your revision
  1. Mark changes in the story

    Split a problem whenever a force or acceleration changes. Label the starting state of each stage before choosing an equation.

  2. Test an inequality before using a friction value

    For an object initially at rest, calculate the friction needed for equilibrium and compare it with the maximum. Do not assume sliding.

  3. Check both translation and rotation

    For static equilibrium, test net force and net torque. A convenient pivot removes one unknown but does not remove the requirement to balance forces.

Send 203-AS1-MS, your assessment topics and an attempted diagram. A tutor can help you identify whether the first difficulty is the physical model, signs or the equations.

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