NEB Class 11 • Physics • Worked Practice

Dynamics Practice Set: NEB 11 Physics

Test Newton’s laws through free-body diagrams, component equations and complete solutions that check systems, directions, units and limiting cases.

Foundation, diagram, friction and multi-body levels
Twelve worked questions plus a timed challenge
Every answer names the system and physical check

Practice boundary

What this Dynamics set measures

The set covers Newton’s laws, system boundaries, free-body diagrams, horizontal forces, equilibrium, friction, inclines and connected bodies. Review the Dynamics concept guide or Dynamics study guide when a setup is unfamiliar. The current CDC Physics Grade 11 page is the Nepal scope checkpoint.

The worked method follows the system-boundary and net-external-force reasoning in OpenStax Newton’s second law. Use current Nepal course notation and depth.

Level 1

Concept questions with explanations

1. Constant velocity

A bus moves east at constant velocity. Must an eastward net force act?

Solution: No. Constant velocity means acceleration is zero, so net external force is zero. Driving force may balance resistance; individual forces need not be zero.

2. Third-law pair

A book rests on a table. Are weight and normal reaction an action–reaction pair?

Solution: No. Both act on the book. The partner of table-on-book normal is book-on-table force. The partner of Earth-on-book weight is book-on-Earth gravitational force.

3. Direction of net force

A ball moves upward after release. What is the net-force direction when air resistance is neglected?

Solution: Downward due to gravity, even while velocity is upward. The opposite directions make the upward speed decrease.

Level 2

Horizontal force and equilibrium problems

4. Opposing forces

A 7 kg trolley has 30 N east and 9 N west. Find acceleration.

Solution: Net force = 21 N east; a = 21/7 = 3.0 m s⁻² east. Horizontal vertical forces cancel if there is no vertical acceleration.

5. Find the applied force

A 12 kg crate accelerates right at 2 m s⁻² while friction is 10 N left. Find the rightward applied force.

Solution: F − 10 = 12×2 = 24, so F = 34 N right. Check: applied force exceeds friction by the required net 24 N.

6. Symmetric cable support

A 200 N sign hangs from two identical cables at 30° above horizontal. Find each tension.

Solution: Horizontal components cancel. Vertical balance: 2T sin30° = 200, so T = 200 N. Each vertical component is 100 N.

Check: More horizontal cables would require greater tension for the same vertical support.

Level 3

Static and kinetic friction problems

7. Static friction adjusts

A 5 N horizontal push acts on a block whose maximum static friction is 12 N. Find friction and acceleration.

Solution: Static friction is 5 N opposite the push, giving zero net force and zero acceleration. It does not automatically take its maximum value.

8. Sliding on a horizontal floor

A 10 kg block slides with μ_k = 0.20. A 40 N horizontal force pulls it. Use g = 9.8 m s⁻².

Solution: N = mg = 98 N. f_k = 0.20×98 = 19.6 N. Net force = 20.4 N; a = 2.04 m s⁻².

9. Angled pull changes normal force

The same block is pulled by 40 N at 30° above horizontal.

Solution: N = 98 − 40 sin30° = 78 N. Friction = 15.6 N. Horizontal pull = 34.6 N. Net = 19.0 N and a ≈ 1.90 m s⁻².

Learning point: Use vertical balance before friction; N is not automatically mg.

Level 4

Inclined-plane questions

10. Frictionless incline

A block slides down a 30° frictionless incline. Find acceleration.

Solution: Down-plane weight component is mg sin30°. Thus a = g sin30° = 4.9 m s⁻². Mass cancels.

Check: As θ approaches zero acceleration approaches zero; as θ approaches 90° it approaches g.

11. Rough incline

A 4 kg block slides down a 25° incline with μ_k = 0.15. Find acceleration using g = 9.8 m s⁻².

Solution: N = mg cos25°. Down-plane net = mg sin25° − μmg cos25°. Therefore a = g(sin25° − 0.15 cos25°) ≈ 2.81 m s⁻² down the plane.

Level 5

Connected-body problem

12. Two blocks on a smooth surface

Blocks of 3 kg and 5 kg are connected by a light string. A 32 N force pulls the 5 kg block. Find acceleration and tension.

Solution: For both blocks as one system, a = 32/(3+5) = 4 m s⁻². For the 3 kg block alone, T = ma = 12 N. For the 5 kg block, 32−12 = 20 N = 5×4 N.

Check: Tension is internal to the combined system and cancels there, but it is external to either individual block.

Use the PhET Forces and Motion simulation after predicting how acceleration changes with force, mass or friction.

Method marks

How to audit a worked Dynamics solution

Give yourself one checkpoint for each reasoning layer: named system, complete external-force inventory, useful axes, correct component equations, algebra with units and a physical check. If the final number is wrong but the diagram and equations are sound, the repair is arithmetic. If the number is correct but the diagram mixes bodies, the method is unsafe and should not receive full self-credit.

Layer	Question to ask	Independent check
System	Which mass does ma refer to?	Do internal forces cancel only for that boundary?
Forces	Which external agent produces each arrow?	Can every arrow be named “A on B”?
Components	Are signs tied to declared axes?	Does the net-force direction match acceleration?
Model	Static, sliding, equilibrium or accelerating?	Are the friction and normal relations conditional?

For a transfer drill, change one feature of Problem 12: apply the same 32 N force to the 3 kg block instead. The common acceleration remains 4 m s⁻², but the tension changes because it must now accelerate the 5 kg block. This shows why a memorised tension formula cannot replace separate diagrams.

15-minute challenge

Mixed questions without method labels

A 15 kg cart accelerates at 1.2 m s⁻² while resistance is 7 N. Find the applied force.
Explain why an action–reaction pair cannot be drawn on one isolated body.
A 50 N push acts 20° below horizontal on a rough floor. Predict how N and friction compare with a horizontal push.
Find acceleration down a 40° frictionless incline.
Two connected blocks have total mass 10 kg and net external force 25 N. Find common acceleration.

Answers: (1) 25 N. (2) the forces act on different bodies. (3) downward component increases N and friction. (4) g sin40°. (5) 2.5 m s⁻².

Error analysis

What to inspect when a Dynamics answer fails

System error

Did you mix forces acting on different bodies in one equation?

Inventory error

Can you name the external agent for every force arrow?

Model error

Did you apply kinetic or limiting-static friction under the right condition?

Direction error

Does the acceleration follow net force, and was the friction tendency predicted?

For online or physical NEB tuition, call 9846662070 with your attempted diagram and first wrong line.

Frequently asked questions

Questions about the Dynamics practice set

How should I score a Dynamics solution?

Credit the system, external-force diagram, axes, component equations, algebra, units and a physical check—not only the final acceleration.

Should I draw a free-body diagram for every question?

For learning and multi-force problems, yes. It exposes the system and directions before algebra hides an error.

Why is normal force not always mg?

Normal force follows perpendicular balance. Angled forces, inclines or perpendicular acceleration can change it.

When does static friction equal μsN?

Only at impending slip in the ideal model. Before that, static friction adjusts to the required value up to the limit.

How do I check connected-body tension?

Substitute the common acceleration into a separate free-body equation for each block. Both should give the same tension.

Where can I get help with NEB Class 11 Dynamics?

For current online or physical tuition options, call 9846662070 and confirm schedule, class mode, teacher availability and fees.

References and next steps

Sources and related study guides

Continue with the Work, Energy & Power practice set to compare force-based and energy-based solutions. Academic scope and sources were checked on 2 August 2026; follow current CDC, NEB and college instructions if requirements change.

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