Figluma mechanics worksheet

Table block and hanging mass

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.

MechanicsConnected particlesFrictionNewton's laws

Diagram State

Rough horizontal table with a block connected over a frictionless pulley to a hanging mass.

table block
hanging mass
light string
frictionless pulley
normal force
kinetic friction
weight forces
tension forces
linked acceleration arrows

Givens

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

Unknowns

Symbol	Quantity	Value	Unit
$a$	Acceleration magnitude	$?$	m s^-2
$T$	String tension	$?$	N

Coordinate System / Sign Convention

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 / Constraints Checklist

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.
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.

Student Solve Checklist

Mark each row only after your setup matches the diagram state; worked equations stay in the teacher key.

SetupAxes, sign convention, model constraints, and linked-motion/origin choices are stated.
Choose linked positive directionsUse the string constraint
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
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
ResultThe final rearrangement, numeric value, units, and direction/speed interpretation are correct.
Eliminate tension and solve for accelerationSubstitute acceleration to find tension

Student Working Area

Solution / Answer Key

1. Choose linked positive directions
$m_1: +x right; m_2: +y downward$
Choose 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.
Equation-choice checkWhat feature of the diagram, sign convention, or givens makes "Choose linked positive directions" the right next equation?Listen for: Choose 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.Flag if: Student can only quote "m_1: +x right; m_2: +y downward" without connecting it to the diagram state or givens.
2. Balance vertical forces on the table block
$N = m_{1} g$
The table block has no vertical acceleration. Its normal force balances its weight, so the friction model can use N = m_1 g.
Equation-choice checkWhat feature of the diagram, sign convention, or givens makes "Balance vertical forces on the table block" the right next equation?Listen for: The table block has no vertical acceleration. Its normal force balances its weight, so the friction model can use N = m_1 g.Flag if: Student can only quote "N = m_1 g" without connecting it to the diagram state or givens.
3. Write kinetic friction on the table block
$f_{k} = μ_{k} N = μ_{k} m_{1} g$
Kinetic friction opposes the table block's rightward motion, so it points left and has magnitude mu_k m_1 g.
Equation-choice checkWhat feature of the diagram, sign convention, or givens makes "Write kinetic friction on the table block" the right next equation?Listen for: Kinetic friction opposes the table block's rightward motion, so it points left and has magnitude mu_k m_1 g.Flag if: Adding kinetic friction to m_2 g instead of subtracting it from the driving force.; Putting friction on the hanging mass or using mu_k m_2 g for the table friction.
4. Use the string constraint
$a_{1} = a_{2} = a$
A light inextensible string over a fixed pulley makes the block's rightward acceleration equal in magnitude to the hanging mass's downward acceleration.
Equation-choice checkWhat feature of the diagram, sign convention, or givens makes "Use the string constraint" the right next equation?Listen for: A light inextensible string over a fixed pulley makes the block's rightward acceleration equal in magnitude to the hanging mass's downward acceleration.Flag if: Treating the two masses as if they could have different accelerations.
5. Write Newton's second law for each body
$T - f_{k} = m_{1} a m_{2} g - T = m_{2} a$
For m_1, tension pulls right and friction acts left. For m_2, weight pulls down and tension acts up. The same tension appears in both equations.
Equation-choice checkWhat feature of the diagram, sign convention, or givens makes "Write Newton's second law for each body" the right next equation?Listen for: For m_1, tension pulls right and friction acts left. For m_2, weight pulls down and tension acts up. The same tension appears in both equations.Flag if: Student can only quote "T - f_k = m_1 a; m_2 g - T = m_2 a" without connecting it to the diagram state or givens.
6. Eliminate tension and solve for acceleration
$a = (m_{2} g - μ_{k} m_{1} g) / (m_{1} + m_{2}) = 2.6 m s^{- 2}$
Adding the two body equations cancels tension. The net external driving force is the hanging weight minus table friction, and the total inertia is m_1 + m_2.
Equation-choice checkWhat feature of the diagram, sign convention, or givens makes "Eliminate tension and solve for acceleration" the right next equation?Listen for: Adding the two body equations cancels tension. The net external driving force is the hanging weight minus table friction, and the total inertia is m_1 + m_2.Flag if: Student can only quote "a = (m_2 g - mu_k m_1 g) / (m_1 + m_2) = 2.6 m s^-2" without connecting it to the diagram state or givens.
7. Substitute acceleration to find tension
$T = m_{1} a + μ_{k} m_{1} g = 18 N$
Use the table block equation after finding acceleration. Tension must both accelerate m_1 and overcome kinetic friction, giving about 18 N.
Equation-choice checkWhat feature of the diagram, sign convention, or givens makes "Substitute acceleration to find tension" the right next equation?Listen for: Use the table block equation after finding acceleration. Tension must both accelerate m_1 and overcome kinetic friction, giving about 18 N.Flag if: Using only m_2 in the denominator after eliminating tension instead of the total mass m_1 + m_2.; Using m_2 g as the tension before accounting for the hanging mass's acceleration.

Diagnostic Checklist

Key checkpoint equations

1. Choose linked positive directions
$m_1: +x right; m_2: +y downward$
2. Balance vertical forces on the table block
$N = m_{1} g$
3. Write kinetic friction on the table block
$f_{k} = μ_{k} N = μ_{k} m_{1} g$
4. Use the string constraint
$a_{1} = a_{2} = a$
5. Write Newton's second law for each body
$T - f_{k} = m_{1} a m_{2} g - T = m_{2} a$
6. Eliminate tension and solve for acceleration
$a = (m_{2} g - μ_{k} m_{1} g) / (m_{1} + m_{2}) = 2.6 m s^{- 2}$
7. Substitute acceleration to find tension
$T = m_{1} a + μ_{k} m_{1} g = 18 N$

Common Wrong Paths

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.

Wrong Answer Signals

Treating the two masses as if they could have different accelerations.Usually indicates the "Use the string constraint" checkpoint needs review.
$a_{1} = a_{2} = a$
Using only m_2 in the denominator after eliminating tension instead of the total mass m_1 + m_2.Usually indicates the "Substitute acceleration to find tension" checkpoint needs review.
$T = m_{1} a + μ_{k} m_{1} g = 18 N$
Adding kinetic friction to m_2 g instead of subtracting it from the driving force.Usually indicates the "Write kinetic friction on the table block" checkpoint needs review.
$f_{k} = μ_{k} N = μ_{k} m_{1} g$
Using m_2 g as the tension before accounting for the hanging mass's acceleration.Usually indicates the "Substitute acceleration to find tension" checkpoint needs review.
$T = m_{1} a + μ_{k} m_{1} g = 18 N$
Putting friction on the hanging mass or using mu_k m_2 g for the table friction.Usually indicates the "Write kinetic friction on the table block" checkpoint needs review.
$f_{k} = μ_{k} N = μ_{k} m_{1} g$
Forgetting that the same tension appears with opposite signs in the two body equations.Usually indicates the "Substitute acceleration to find tension" checkpoint needs review.
$T = m_{1} a + μ_{k} m_{1} g = 18 N$

Tutor Marking Rubric

Tutor score rows use curated Figluma checkpoints as marking cues. They are not automated grading or a symbolic mark scheme.

Tutor Mark Sheet

Manual tutor mark sheet only. Use observed work and leave rows blank when evidence is copied from a reveal.

Total: ___ / 8

Setup

Choose linked positive directions; Use the string constraint

___ / 2

Axes, sign convention, model constraints, and linked-motion/origin choices are stated.If weak, reteach the row from this signal: Treating the two masses as if they could have different accelerations.

Score descriptions

0No usable evidence for this row, or the work contradicts "Choose linked positive directions".
1Partly correct, but review this row's checkpoint signal: Treating the two masses as if they could have different accelerations.
2Complete row: Axes, sign convention, model constraints, and linked-motion/origin choices are stated.

Watch signals

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

Components

Balance vertical forces on the table block; Write kinetic friction on the table block

___ / 2

Resolved components, force directions, normal/friction setup, or velocity split are correct.If weak, reteach the row from this signal: Adding kinetic friction to m_2 g instead of subtracting it from the driving force.

Score descriptions

0No usable evidence for this row, or the work contradicts "Balance vertical forces on the table block".
1Partly correct, but review this row's checkpoint signal: Adding kinetic friction to m_2 g instead of subtracting it from the driving force.
2Complete row: Resolved components, force directions, normal/friction setup, or velocity split are correct.

Watch signals

Adding kinetic friction to m_2 g instead of subtracting it from the driving force.
Putting friction on the hanging mass or using mu_k m_2 g for the table friction.

Net-force / governing equation

Write Newton's second law for each body

___ / 2

The main Newton's law or motion equation uses the right model, signs, and shared variables.If weak, reteach from "Write Newton's second law for each body" before assigning another variant.

Score descriptions

0No usable evidence for this row, or the work contradicts "Write Newton's second law for each body".
1Partly correct, but review this row's checkpoint signal: evidence reaches "Write Newton's second law for each body" but is not yet consistent across the row
2Complete row: The main Newton's law or motion equation uses the right model, signs, and shared variables.

Watch signals

Use the checkpoint titles when no curated wrong-path signal is listed for this row.

Result

Eliminate tension and solve for acceleration; Substitute acceleration to find tension

___ / 2

The final rearrangement, numeric value, units, and direction/speed interpretation are correct.If weak, reteach the row from this signal: Using only m_2 in the denominator after eliminating tension instead of the total mass m_1 + m_2.

Score descriptions

0No usable evidence for this row, or the work contradicts "Eliminate tension and solve for acceleration".
1Partly correct, but review this row's checkpoint signal: Using only m_2 in the denominator after eliminating tension instead of the total mass m_1 + m_2.
2Complete row: The final rearrangement, numeric value, units, and direction/speed interpretation are correct.

Watch signals

Using only m_2 in the denominator after eliminating tension instead of the total mass m_1 + m_2.
Using m_2 g as the tension before accounting for the hanging mass's acceleration.

Total: ___ / 8Add the four row scores after checking actual student evidence. Leave it blank when the work is incomplete or only copied from reveals.Reteach from the first 0 or 1 row before assigning another variant or adding a new problem.

Setup

Axes, sign convention, model constraints, and linked-motion/origin choices are stated.

012

Score guide

0No usable evidence for this row, or the work contradicts "Choose linked positive directions".
1Partly correct, but review this row's checkpoint signal: Treating the two masses as if they could have different accelerations.
2Complete row: Axes, sign convention, model constraints, and linked-motion/origin choices are stated.

Checkpoints

Choose linked positive directions
$m_1: +x right; m_2: +y downward$
Use the string constraint
$a_{1} = a_{2} = a$

Watch for

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

Components

Resolved components, force directions, normal/friction setup, or velocity split are correct.

012

Score guide

0No usable evidence for this row, or the work contradicts "Balance vertical forces on the table block".
1Partly correct, but review this row's checkpoint signal: Adding kinetic friction to m_2 g instead of subtracting it from the driving force.
2Complete row: Resolved components, force directions, normal/friction setup, or velocity split are correct.

Checkpoints

Balance vertical forces on the table block
$N = m_{1} g$
Write kinetic friction on the table block
$f_{k} = μ_{k} N = μ_{k} m_{1} g$

Watch for

Adding kinetic friction to m_2 g instead of subtracting it from the driving force.
Putting friction on the hanging mass or using mu_k m_2 g for the table friction.

Net-force / governing equation

The main Newton's law or motion equation uses the right model, signs, and shared variables.

012

Score guide

0No usable evidence for this row, or the work contradicts "Write Newton's second law for each body".
1Partly correct, but review this row's checkpoint signal: evidence reaches "Write Newton's second law for each body" but is not yet consistent across the row
2Complete row: The main Newton's law or motion equation uses the right model, signs, and shared variables.

Checkpoints

Write Newton's second law for each body
$T - f_{k} = m_{1} a m_{2} g - T = m_{2} a$

Watch for

Use the checkpoint equations for this row.

Result

The final rearrangement, numeric value, units, and direction/speed interpretation are correct.

012

Score guide

0No usable evidence for this row, or the work contradicts "Eliminate tension and solve for acceleration".
1Partly correct, but review this row's checkpoint signal: Using only m_2 in the denominator after eliminating tension instead of the total mass m_1 + m_2.
2Complete row: The final rearrangement, numeric value, units, and direction/speed interpretation are correct.

Checkpoints

Eliminate tension and solve for acceleration
$a = (m_{2} g - μ_{k} m_{1} g) / (m_{1} + m_{2}) = 2.6 m s^{- 2}$
Substitute acceleration to find tension
$T = m_{1} a + μ_{k} m_{1} g = 18 N$

Watch for

Using only m_2 in the denominator after eliminating tension instead of the total mass m_1 + m_2.
Using m_2 g as the tension before accounting for the hanging mass's acceleration.
Forgetting that the same tension appears with opposite signs in the two body equations.

Diagram State

Givens

Unknowns

Coordinate System / Sign Convention

Assumptions / Constraints Checklist

Student Solve Checklist

Student Working Area

1. Choose linked positive directions

2. Balance vertical forces on the table block

3. Write kinetic friction on the table block

4. Use the string constraint

5. Write Newton's second law for each body

6. Eliminate tension and solve for acceleration

7. Substitute acceleration to find tension

Diagnostic Checklist

Key checkpoint equations

Common Wrong Paths

Wrong Answer Signals

Tutor Marking Rubric

Tutor Mark Sheet

Setup

Components

Net-force / governing equation

Result

Setup

Score guide

Checkpoints

Watch for

Components

Score guide

Checkpoints

Watch for

Net-force / governing equation

Score guide

Checkpoints

Watch for

Result

Score guide

Checkpoints

Watch for