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He applied a force to the box and it moved. Force applied on it equals to the weight. As he lifts the point of application of the force gets displaced. Hence work is said to be done.
potential energy
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While lifting it straight up is harder, overall it is less work because you do not have to contend with the friction of the inclined plane. Overcoming that friction is work done which is totally wasted.While lifting it straight up is harder, overall it is less work because you do not have to contend with the friction of the inclined plane. Overcoming that friction is work done which is totally wasted.While lifting it straight up is harder, overall it is less work because you do not have to contend with the friction of the inclined plane. Overcoming that friction is work done which is totally wasted.While lifting it straight up is harder, overall it is less work because you do not have to contend with the friction of the inclined plane. Overcoming that friction is work done which is totally wasted.
Work is done in lifting is equal to work done in opposing gravity. Work done in opposing gravity = weight times distance = 700 x 4 =2800 joules
4
0 J
The same type as is done when pushing, shoving, rolling, lifting, wiggling etc. ... a force acting through a distance.
The work done by lifting two loads up one story vs the work done lifting one load will depend on the weight of the second load. If it is the same as the first load, it would simply be twice the work to lift. Work is described as a force times a distance and a force is described as a mass times an acceleration. In this case, the force would be twice as much, so the work would be as well.
Work = energy (by the Work-Energy Equivalence Theorem).The work done in lifting an object shows up as the gravitational potential energy of the object: W = mgh, where W is the work done; m is the object's mass; g is the acceleration due to Earth's gravity; and h is the height to which the object is lifted. (This assumes that the lifting is done near Earth's surface, with gravity effectively constant.)ALL Credit goes to Argent at yahoo answers
That's correct. For work to be done, the force must have a component in the direction of the movement.
The answer is 0. If you used the formula of =>mgh, you would get a value which would be wrong. The question is a trick question. There is no work done in 'holding' an object but work would be done in 'lifting' an object. Hence, 0. ~Dartz