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Motion, Forces and Energy - Forces (Resultant forces, Hooke's Law, and circular motion)

Grade 11IGCSEPhysics

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

🔑Concepts

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A resultant force FF causes an object of mass mm to accelerate with acceleration aa in the direction of the force, described by Newton's Second Law: F=maF = ma.

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If the resultant force on an object is zero (Fnet=0F_{net} = 0), the object will remain at rest or continue to move at a constant velocity.

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Hooke's Law states that the extension xx of a spring is directly proportional to the applied load FF, provided the limit of proportionality is not exceeded (F=kxF = kx).

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The spring constant kk is a measure of the stiffness of the spring, measured in N/mN/m or N/cmN/cm.

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In circular motion, an object moving at a constant speed is still accelerating because its direction is constantly changing. This is called centripetal acceleration.

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Centripetal force FcF_c is the resultant force acting towards the center of the circle required to keep an object in circular motion. It depends on mass mm, velocity vv, and radius rr.

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Friction, gravity, or tension can act as the centripetal force depending on the scenario (e.g., a car turning, a planet orbiting, or a stone on a string).

📐Formulae

F=maF = ma

F=kxF = kx

x=lfinal−loriginalx = l_{final} - l_{original}

Fc=mv2rF_c = \frac{mv^2}{r}

Resultant Force (Fres)=Fforward−Ffriction\text{Resultant Force } (F_{res}) = F_{forward} - F_{friction}

💡Examples

Problem 1:

A spring has an original length of 12.0 cm12.0\text{ cm}. When a weight of 6.0 N6.0\text{ N} is hung from it, the new length is 15.0 cm15.0\text{ cm}. Calculate the spring constant kk in N/mN/m.

Solution:

x=15.0 cm−12.0 cm=3.0 cm=0.03 mx = 15.0\text{ cm} - 12.0\text{ cm} = 3.0\text{ cm} = 0.03\text{ m}. Using F=kxF = kx, k=Fx=6.00.03=200 N/mk = \frac{F}{x} = \frac{6.0}{0.03} = 200\text{ N/m}.

Explanation:

First, find the extension by subtracting the original length from the final length. Convert the extension to meters to find the spring constant in standard SI units (N/mN/m).

Problem 2:

A car of mass 800 kg800\text{ kg} is traveling around a circular track of radius 50 m50\text{ m} at a constant speed of 20 m/s20\text{ m/s}. Calculate the centripetal force required to keep the car on the track.

Solution:

Fc=mv2r=800×(20)250=800×40050=6400 NF_c = \frac{mv^2}{r} = \frac{800 \times (20)^2}{50} = \frac{800 \times 400}{50} = 6400\text{ N}.

Explanation:

Apply the centripetal force formula using the mass, velocity squared, and the radius of the path.

Problem 3:

An object of mass 5 kg5\text{ kg} is pulled to the right with a force of 25 N25\text{ N} and to the left with a frictional force of 5 N5\text{ N}. Determine the acceleration of the object.

Solution:

Fres=25 N−5 N=20 NF_{res} = 25\text{ N} - 5\text{ N} = 20\text{ N}. Using F=maF = ma, a=20 N5 kg=4 m/s2a = \frac{20\text{ N}}{5\text{ kg}} = 4\text{ m/s}^2.

Explanation:

Find the resultant force by subtracting the opposing forces, then use Newton's Second Law to solve for acceleration.

Forces (Resultant forces, Hooke's Law, and circular motion) Revision - Grade 11 Physics IGCSE