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Block on a rough inclineForces / Inclined planes / Friction Crate pulled across a rough floorForces / Horizontal motion / Friction Table block and hanging massConnected particles / Friction / Newton's laws Passenger in an accelerating elevatorNewton's laws / Apparent weight / Vertical motion Projectile launched from a cliffProjectile motion / Kinematics / Vectors

A 4.0 kg block on a rough horizontal table is connected by a light string over a frictionless pulley to a 2.5 kg hanging mass. The coefficient of kinetic friction between the block and table is 0.20. Find the acceleration of the system and the tension in the string.

State

Givens

$m_{1}$ Mass on table: $4.0 kg$
$m_{2}$ Hanging mass: $2.5 kg$
$μ_{k}$ Kinetic friction coefficient: $0.20$
$g$ Gravitational field strength: $9.8 m s^{- 2}$

Unknowns

$a$ Acceleration magnitude: $? m s^{- 2}$
$T$ String tension: $? N$

For m_1, +x is right toward the pulley. For m_2, +y is downward, in the direction of the hanging mass's motion.

Assumptions

The hanging mass descends and the table block slides right, so kinetic friction on m_1 points left.
The string is light and inextensible, so both masses have the same acceleration magnitude.
The pulley is frictionless, so the string tension has the same magnitude on both sides.
The table block stays on the table, so its vertical acceleration is zero.
Air resistance is negligible.

Constraints

The two bodies must share one acceleration magnitude because they are joined by one taut string.
The tension is internal to the two-body system, so it cancels when the two Newton's law equations are added.
Kinetic friction on m_1 is f_k = mu_k m_1 g because N = m_1 g on the horizontal table.
The model assumes m_2 g is greater than kinetic friction, so the stated direction of motion is consistent.

Table block and hanging mass

Active checkpoint: Choose linked positive directions

LookSetupChoose linked positive directions

ChooseChoose linked positive directionsChoose positive directions along the expected motion for both bodies. The table block moves right and the hanging mass moves downward, so both accelerations can be written as +a.

AvoidWrong turnStudent can only quote "m_1: +x right; m_2: +y downward" without connecting it to the diagram state or givens.

Equation lens

T a r g e t

m_1: +x right; m_2: +y downward

Choose linked positive directionsBefore revealing the worked equation, decide what "Choose linked positive directions" must establish from the diagram and givens.

Hint before equation

Set the axes, origin, sign convention, or linked-motion choice before writing algebra.

Treating the two masses as if they could have different accelerations.

Equation-choice checkWhat feature of the diagram, sign convention, or givens makes "Choose linked positive directions" the right next equation?

Step 1 of 7SetupChoose linked positive directions

Use the diagram focus, problem state, and scratch workspace first.

Checkpoints

Choose linked positive directions
Not yet
Set the axes, origin, sign convention, or linked-motion choice before writing algebra.
Watch: Treating the two masses as if they could have different accelerations.
Balance vertical forces on the table block
Not yet
Resolve only onto the chosen axes and check which part of each vector belongs in this step.
Watch: Adding kinetic friction to m_2 g instead of subtracting it from the driving force.
Write kinetic friction on the table block
Not yet
Separate driving terms from opposing terms, then assign signs from the positive direction.
Watch: Adding kinetic friction to m_2 g instead of subtracting it from the driving force.
Use the string constraint
Not yet
Set the axes, origin, sign convention, or linked-motion choice before writing algebra.
Watch: Treating the two masses as if they could have different accelerations.
Write Newton's second law for each body
Not yet
Separate driving terms from opposing terms, then assign signs from the positive direction.
Eliminate tension and solve for acceleration
Not yet
Substitute after the model equation is settled, then check units and the requested interpretation.
Watch: Using only m_2 in the denominator after eliminating tension instead of the total mass m_1 + m_2.
Substitute acceleration to find tension
Not yet
Substitute after the model equation is settled, then check units and the requested interpretation.
Watch: Using only m_2 in the denominator after eliminating tension instead of the total mass m_1 + m_2.

Common traps

Treating the two masses as if they could have different accelerations.
Using only m_2 in the denominator after eliminating tension instead of the total mass m_1 + m_2.
Adding kinetic friction to m_2 g instead of subtracting it from the driving force.
Using m_2 g as the tension before accounting for the hanging mass's acceleration.
Putting friction on the hanging mass or using mu_k m_2 g for the table friction.
Forgetting that the same tension appears with opposite signs in the two body equations.

Solve Workspace

Solve checklist0 / 4 marked

Checkpoint reveal0 / 7 revealed

Checkpoint progress0 / 7 worked; 0 checked

Equation reasons0 / 7 noted

SetupAxes, sign convention, model constraints, and linked-motion/origin choices are stated.
Choose linked positive directionsUse the string constraint
1 common trap tied to this row.
ComponentsResolved components, force directions, normal/friction setup, or velocity split are correct.
Balance vertical forces on the table blockWrite kinetic friction on the table block
2 common traps tied to this row.
Net-force / governing equationThe main Newton's law or motion equation uses the right model, signs, and shared variables.
Write Newton's second law for each body
Use the checkpoint cues before moving on.
ResultThe final rearrangement, numeric value, units, and direction/speed interpretation are correct.
Eliminate tension and solve for accelerationSubstitute acceleration to find tension
3 common traps tied to this row.

Why this equation?

Local note for the active checkpoint. It is copied into the attempt snapshot for validation.

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