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Physics - Force, Work, Power and Energy

Grade 10ICSE

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

🔑Concepts

Moment of Force (Torque): The turning effect of a force about a fixed point or axis. It is calculated as the product of the force and the perpendicular distance from the axis of rotation: τ=F×d\tau = F \times d.

Principle of Moments: For a body in equilibrium, the sum of clockwise moments about a pivot is equal to the sum of anticlockwise moments about the same pivot: Clockwise Moments=Anticlockwise Moments\sum \text{Clockwise Moments} = \sum \text{Anticlockwise Moments}.

Work: Work is said to be done only when a force applied on a body makes the body move. It is a scalar quantity defined as W=FscosθW = Fs \cos \theta, where θ\theta is the angle between the force and displacement.

Power: The rate of doing work. It is the ratio of work done to the time taken: P=WtP = \frac{W}{t}. Its S.I. unit is the Watt (1 W=1 J/s1 \text{ W} = 1 \text{ J/s}).

Energy: The capacity to do work. Mechanical energy exists in two forms: Potential Energy (UU), due to position or configuration, and Kinetic Energy (KK), due to motion.

Work-Energy Theorem: The work done by a force on a moving body is equal to the increase in its kinetic energy: W=ΔK=12mv212mu2W = \Delta K = \frac{1}{2}mv^2 - \frac{1}{2}mu^2.

Law of Conservation of Energy: Energy can neither be created nor destroyed; it can only be transformed from one form to another. In a conservative field (like gravity), the total mechanical energy (K+UK + U) remains constant.

Units of Energy: Common units include the Joule (JJ), Calorie (1 cal=4.18 J1 \text{ cal} = 4.18 \text{ J}), Electron-volt (1 eV=1.6×1019 J1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}), and Kilowatt-hour (1 kWh=3.6×106 J1 \text{ kWh} = 3.6 \times 10^6 \text{ J}).

📐Formulae

Moment of Force (Torque)=F×d\text{Moment of Force (Torque)} = F \times d

W=FscosθW = Fs \cos \theta

P=Wt=F×vP = \frac{W}{t} = F \times v

U=mghU = mgh

K=12mv2K = \frac{1}{2}mv^2

K=p22m (where p is linear momentum)K = \frac{p^2}{2m} \text{ (where } p \text{ is linear momentum)}

Etotal=K+U=constant (during free fall)E_{\text{total}} = K + U = \text{constant (during free fall)}

W=12m(v2u2)W = \frac{1}{2}m(v^2 - u^2)

💡Examples

Problem 1:

A body of mass 5 kg5 \text{ kg} is taken from a height of 5 m5 \text{ m} to 10 m10 \text{ m}. Find the increase in its potential energy. (Take g=10 m/s2g = 10 \text{ m/s}^2)

Solution:

ΔU=mgh2mgh1=mg(h2h1)\Delta U = mgh_2 - mgh_1 = mg(h_2 - h_1). Substituting values: ΔU=5×10×(105)=50×5=250 J\Delta U = 5 \times 10 \times (10 - 5) = 50 \times 5 = 250 \text{ J}.

Explanation:

The change in potential energy depends on the change in vertical height above the reference level.

Problem 2:

A force of 10 N10 \text{ N} acts on a body and displaces it by 2 m2 \text{ m} in a direction making an angle of 6060^\circ with the direction of the force. Calculate the work done.

Solution:

W=Fscosθ=10×2×cos60=20×0.5=10 JW = Fs \cos \theta = 10 \times 2 \times \cos 60^\circ = 20 \times 0.5 = 10 \text{ J}.

Explanation:

Work done is calculated using the component of force in the direction of displacement.

Problem 3:

A ball of mass 0.5 kg0.5 \text{ kg} is dropped from a height of 20 m20 \text{ m}. Calculate its velocity just before hitting the ground. (Ignore air resistance, g=10 m/s2g = 10 \text{ m/s}^2)

Solution:

By Conservation of Energy: Utop=Kbottommgh=12mv2U_{\text{top}} = K_{\text{bottom}} \Rightarrow mgh = \frac{1}{2}mv^2. Solving for vv: v=2gh=2×10×20=400=20 m/sv = \sqrt{2gh} = \sqrt{2 \times 10 \times 20} = \sqrt{400} = 20 \text{ m/s}.

Explanation:

The initial gravitational potential energy at the height is entirely converted into kinetic energy at the ground level.

Force, Work, Power and Energy - Revision Notes & Key Formulas | ICSE Class 10 Science