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A porter lifts a 10 kg luggage from the ground and places it on his head at a height of 2.5 m. What is the work done by the porter on the luggage? (Using g =10 m/s 2 )
25 J
500 J
100 J
250 J
250 J
The work done by an external agent in lifting an object is equal to the change in its gravitational potential energy. Given the mass m=10┬аkg, height h=2.5┬аm, and gravity g=10┬аm/s2, the work done is W=mgh=10├Ч10├Ч2.5=250┬аJ.
The work done by an external agent in lifting an object is equal to the change in its gravitational potential energy. Given the mass m=10┬аkg, height h=2.5┬аm, and gravity g=10┬аm/s2, the work done is W=mgh=10├Ч10├Ч2.5=250┬аJ.
Lifting a bag is like charging a battery; you are putting energy into the system that is stored as potential height.
Work is 'm-g-h' (M-G-Height): Mass times Gravity times Height.
W=Fdcos(╬╕) тАФ General work formula where ╬╕ is the angle between force and displacement.
U=mgh тАФ Gravitational potential energy stored in an object at height h.
When an object is moved against the gravitational field, work is done to increase its potential energy. Since the motion is uniform (without change in kinetic energy), the applied force must be equal and opposite to the gravitational force F=mg. Thus, work W=F├Чd=(mg)├Чh.
Work is a scalar quantity measured in Joules (J).
If the object is lifted at a constant velocity, the net force is zero, but the work done by the agent is strictly positive.
The displacement is in the direction of the applied force, hence ╬╕=0┬░ and cos(0┬░)=1.
Useful for calculating energy requirements in mechanical systems.
Applicable to any vertical displacement in a uniform gravitational field.
Does not account for frictional losses or air resistance.
Assumes g is constant over the vertical distance.
Calculating the power consumption of elevators.
Determining the energy expended in construction or manual labor.
The standard acceleration due to gravity on Earth is gтЙИ9.8┬аm/s2, though 10┬аm/s2 is often used in competitive exams for calculation ease.
Option A (25 J) results from missing the gravity factor g.
Option C (100 J) ignores the height component or uses incorrect variables.
D is correct тАФ The work done by the porter is calculated using the formula W=mgh, which equals 10├Ч10├Ч2.5=250┬аJ.
Always ensure the units are SI (kg, m, m/s┬▓) before substituting into formulas to avoid conversion errors.
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