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Work Energy and Machines - Numerical Problems

Grade 8CBSE

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

🔑Concepts

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Work is done when a force FF acts on an object and causes displacement ss in the direction of the force.

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Energy is defined as the capacity to do work. Its SI unit is Joule (JJ).

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Potential Energy (EpE_p) is the energy possessed by a body by virtue of its position or shape.

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Kinetic Energy (EkE_k) is the energy possessed by a body by virtue of its motion.

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Power is the rate of doing work or the rate of transfer of energy. Its SI unit is Watt (WW).

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Mechanical Advantage (MAMA) is the ratio of the Load (LL) to the Effort (EE).

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Velocity Ratio (VRVR) is the ratio of the distance moved by effort (dEd_E) to the distance moved by load (dLd_L).

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Efficiency (η\eta) of a machine is the ratio of useful work output to the work input, often expressed as a percentage.

📐Formulae

W=F×sW = F \times s

Ep=m×g×hE_p = m \times g \times h

Ek=12mv2E_k = \frac{1}{2} m v^2

P=WtP = \frac{W}{t}

MA=Load (L)Effort (E)MA = \frac{\text{Load (L)}}{\text{Effort (E)}}

VR=dEdLVR = \frac{d_E}{d_L}

η=MAVR×100%\eta = \frac{MA}{VR} \times 100\%

💡Examples

Problem 1:

Calculate the work done when a force of 50 N50\text{ N} displaces a stone through a distance of 8 m8\text{ m} in the direction of the force.

Solution:

Given: Force F=50 NF = 50\text{ N} Displacement s=8 ms = 8\text{ m} Formula: W=F×sW = F \times s W=50×8W = 50 \times 8 W=400 JW = 400\text{ J}

Explanation:

Work is the product of force and displacement. Since the displacement is in the direction of the force, we simply multiply the two values.

Problem 2:

A body of mass 5 kg5\text{ kg} is raised to a height of 10 m10\text{ m}. Calculate its potential energy. (Take g=9.8 m/s2g = 9.8\text{ m/s}^2)

Solution:

Given: Mass m=5 kgm = 5\text{ kg} Height h=10 mh = 10\text{ m} Acceleration due to gravity g=9.8 m/s2g = 9.8\text{ m/s}^2 Formula: Ep=m×g×hE_p = m \times g \times h Ep=5×9.8×10E_p = 5 \times 9.8 \times 10 Ep=490 JE_p = 490\text{ J}

Explanation:

Gravitational potential energy depends on the mass, the height above the reference point, and the acceleration due to gravity.

Problem 3:

An electric motor performs 12000 J12000\text{ J} of work in 2 minutes2\text{ minutes}. Find the power of the motor in Watts.

Solution:

Given: Work W=12000 JW = 12000\text{ J} Time t=2 minutes=2×60=120 st = 2\text{ minutes} = 2 \times 60 = 120\text{ s} Formula: P=WtP = \frac{W}{t} P=12000120P = \frac{12000}{120} P=100 WP = 100\text{ W}

Explanation:

Power is work divided by time. It is crucial to convert time into the SI unit (seconds) before calculation.

Problem 4:

A machine overcomes a load of 800 N800\text{ N} by applying an effort of 200 N200\text{ N}. Calculate the Mechanical Advantage (MAMA) of the machine.

Solution:

Given: Load L=800 NL = 800\text{ N} Effort E=200 NE = 200\text{ N} Formula: MA=LEMA = \frac{L}{E} MA=800200MA = \frac{800}{200} MA=4MA = 4

Explanation:

Mechanical Advantage is a ratio of forces, so it has no units. It indicates how many times the machine multiplies the input force.

Problem 5:

An object of mass 2 kg2\text{ kg} is moving with a velocity of 4 m/s4\text{ m/s}. Find its kinetic energy.

Solution:

Given: Mass m=2 kgm = 2\text{ kg} Velocity v=4 m/sv = 4\text{ m/s} Formula: Ek=12mv2E_k = \frac{1}{2} m v^2 Ek=12×2×(4)2E_k = \frac{1}{2} \times 2 \times (4)^2 Ek=1×16E_k = 1 \times 16 Ek=16 JE_k = 16\text{ J}

Explanation:

Kinetic energy is proportional to the mass and the square of the velocity of the object.