Work done by a constant forceAssume you are lifting an object with mass 20 kg from the ground to a height of 1.5 m. Assume that you are exerting a constant force in the upward direction and that you are moving the object upward with uniform velocity. The net force on the object is zero. The force you are exerting is equal in magnitude and opposite in direction to the force of gravity. As you are lifting the object you are doing work on the object. Show
The work W done on an object by a constant force is defined as W = F·d. It is equal to the magnitude of the force, multiplied by the distance the object moves in the direction of the force. Work is a scalar, a number with units. The SI unit of work Nm = Joule (J). Work is the "scalar product" or "dot product" of the force and the displacement vector. The scalar product of two vectors A and B is a scalar quantity (a number with units) equal to the product of the magnitudes of the two vectors and the cosine of the smallest angle between them. A·B = ABcosθ. In terms of the Cartesian components of the vectors A and B the scalar product is written as A·B = AxBx + AyBy + AzBz.
Example:Assume you forgot to set the parking break and your car starts rolling down a hill. You try in vain to stop it by pulling as hard as you can on the bumper, but the car keeps on moving forward. You exert a force on the car opposite to the direction of travel. The distance traveled in the direction of the force is negative, you do negative work on the car. But the car is pulling you in the direction of travel with a force of equal magnitude (Newton's third law). The car is doing positive work on you. The net force on the object is zero, Fa + Fg = 0. The displacement vector points in the upward direction. The work done by the applied force Fa, W = Fa·d = Fad = mgd, is positive. The work done by the gravitational force Fg, W = Fg·d = -Fgd = -mgd, is negative. The net work done by all forces acting on the object Wnet = Fnet·d is zero. Problem:For A = 3i + j - k, B = -i + 2j + 5k, and C = 2j - 3k, find C·(A - B). Solution:
Problem:Explain why the work done by the force of sliding friction is negative when an object undergoes a displacement on a rough surface? Solution:
Problem:When a particle moves in a circle, a force acts on it, directed towards the center of rotation. Why is it that this force does no work on the particle? Solution:
Problem:While mowing a lawn, a boy pushes a lawn mower a total distance of 350 m over the grass with a force of 90 N directed along the horizontal. How much work is done by the boy? Solution:
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