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Laws of Motion - Solving Problems in Mechanics

Grade 11CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A Free Body Diagram (FBD) is a diagrammatic representation of a single body in isolation, showing all the external forces acting on it represented by vectors.

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The choice of the 'system' is crucial; it can be a single particle, a rigid body, or a collection of bodies, provided all parts have the same acceleration a⃗\vec{a}.

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According to Newton's Second Law, the vector sum of all external forces acting on a body is equal to the product of its mass and acceleration: ∑F⃗=ma⃗\sum \vec{F} = m\vec{a}.

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In most problems, we resolve forces into mutually perpendicular components (usually along the xx and yy axes) such that ∑Fx=max\sum F_x = ma_x and ∑Fy=may\sum F_y = ma_y.

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Normal reaction NN is the contact force exerted by a surface on an object, acting perpendicular to the surface of contact.

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Tension TT is the pulling force transmitted through a string, rope, or chain. For an ideal (massless and inextensible) string passing over a frictionless pulley, the tension is uniform throughout.

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Equilibrium of a particle occurs when the net external force on the particle is zero, i.e., ∑F⃗=0\sum \vec{F} = 0, implying the particle is either at rest or moving with constant velocity.

📐Formulae

F⃗net=ma⃗\vec{F}_{net} = m\vec{a}

fs≤μsNf_s \leq \mu_s N

fk=μkNf_k = \mu_k N

a=Net Pulling ForceTotal Massa = \frac{\text{Net Pulling Force}}{\text{Total Mass}} (for simple connected systems)

T=m(g+a)T = m(g + a) (Object moving upward with acceleration aa)

T=m(g−a)T = m(g - a) (Object moving downward with acceleration aa)

a=m2−m1m1+m2ga = \frac{m_2 - m_1}{m_1 + m_2}g (Atwood machine acceleration where m2>m1m_2 > m_1)

T=2m1m2m1+m2gT = \frac{2m_1 m_2}{m_1 + m_2}g (Tension in Atwood machine)

💡Examples

Problem 1:

Two masses m1=5 kgm_1 = 5 \text{ kg} and m2=10 kgm_2 = 10 \text{ kg} are connected by a light inextensible string passing over a frictionless pulley. If the system is released from rest, find the acceleration of the masses and the tension in the string. (Take g=10 m/s2g = 10 \text{ m/s}^2)

Solution:

Let aa be the acceleration of the system and TT be the tension. For m2m_2 (moving down): m2g−T=m2a  ⟹  100−T=10am_2 g - T = m_2 a \implies 100 - T = 10a For m1m_1 (moving up): T−m1g=m1a  ⟹  T−50=5aT - m_1 g = m_1 a \implies T - 50 = 5a Adding the two equations: (100−T)+(T−50)=10a+5a(100 - T) + (T - 50) = 10a + 5a 50=15a  ⟹  a=5015=103 m/s250 = 15a \implies a = \frac{50}{15} = \frac{10}{3} \text{ m/s}^2 Substituting aa into the second equation: T=50+5(103)=50+503=2003 NT = 50 + 5\left(\frac{10}{3}\right) = 50 + \frac{50}{3} = \frac{200}{3} \text{ N}

Explanation:

Since the string is inextensible, both masses have the same magnitude of acceleration. We treat each mass as a separate system, draw their FBDs, and apply F=maF = ma in the direction of motion.

Problem 2:

A block of mass m=2 kgm = 2 \text{ kg} is placed on a smooth inclined plane making an angle of θ=30∘\theta = 30^\circ with the horizontal. Calculate the acceleration of the block down the plane.

Solution:

The forces acting on the block are gravity mgmg and normal force NN. Resolving mgmg into components parallel and perpendicular to the incline: Force parallel to the incline: F∥=mgsin⁡θF_{\parallel} = mg \sin \theta Force perpendicular to the incline: F⊥=N−mgcos⁡θ=0F_{\perp} = N - mg \cos \theta = 0 Applying Newton's second law along the incline: mgsin⁡θ=mamg \sin \theta = ma a=gsin⁡30∘=10×12=5 m/s2a = g \sin 30^\circ = 10 \times \frac{1}{2} = 5 \text{ m/s}^2

Explanation:

On an inclined plane, the component of gravity acting down the slope is mgsin⁡θmg \sin \theta. Because the surface is smooth (frictionless), this is the only force causing acceleration along the plane.