Work Done by Gravity

WORK DONE BY A CONSTANT FORCE

The work W done by a constant force F when its point of application undergoes a displacement s is defined to be

W = Fs cos θ (4.l a)

where θ is the angle between F and s. Only the component of F along s, that is, F cos θ, contributes to the work done. Strictly speaking, the work is done by the source or agent that applies the force. Work is a scalar quantity and its SI unit is the joule (J). From Eq. 7.1a we see that

1 J = 1 N · m

Equation 4.1 a may be expressed in terms of the dot product:

W = F · s (4.1 b)

In terms of rectangular components, the two vectors are F = F _x i + F _y j + F _z k and s = ∆x i + ∆y j + ∆z k; hence, Eq. 4.1 b may be written as

W = F _x ∆x + F _y ∆y + F _z ∆z (4.1 c)

The work done by a given force on a body depends only on the force, the displacement, and the angle between them. It does not depend on the velocity or the acceleration of the body, or on the presence of other forces. Since work is a scalar, its value also does not depend on the orientation of the coordinate axes. However, since the magnitude of a displacement in a given time interval depends on the velocity of the frame of reference used to measure the displacement, the calculated work also depends on the reference frame.

From Eq. 4.1 a we see that when the force and the displacement are perpen­dicular, no work is done. Thus, when a block slides along a plane the normal force N does no work. Similarly, a centripetal force, such as the tension F in a rope does no work because it is always perpendicular to the motion. Work was initially required to give the object its speed, but none is needed to maintain it. This is a situation in which a force accelerates a particle, yet does no work.

When several forces act on a body one may calculate the work done by each force individually. The net work done by the body is the algebraic sum of the individual contributions:

W _net = F ₁ s ₁ + F ₂ s ₂ + … + F _n s _n

FIGURE 4.1

Figure 4.1 shows a block that undergoes a displacement along an incline. Let us find the work done by the force of gravity on the block. To use Eq. 4.1 c, we need to choose coordinate axes to specify the components of the two vectors − although the value of the work will not depend on the orientation of the axes. With the choice of axes indicated, m g = −mg j. In terms of rectangular components, a finite displacement is always expressed in the form

s = ∆ x i + ∆ y j + ∆ z k

Thus, the work done by gravity on the block is (with ∆ z = 0)

W_g = m g · s = (−mg j ) · ( ∆ x i + ∆ y j)

Since ∆ y = y _f - y _i, we have

W_g = −mg (y _f − y _i) (4.2)

In Fig. 4.1, ∆ y = y _f − y _i = − h, so W_g = +mgh. If the displacement were in the opposite direction, Eq. 4.2 would still have the same form; one would simply substitute the appropriate values for y _f and y _i.

From Eq.4.2 we infer an important result: The work done by the force of gravity depends only on the initial and final vertical coordinates, not on the path taken. Furthermore, if y _f = y _i, then W_g = 0. That is, the work done by gravity is zero for any path that returns to the initial point. These facts will assume great importance in the next chapter.

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