CHAPTER 6. WORK, KINETIC ENERGY AND POTENTIAL ENERGY
The force of a bow string (a) on the object pulling it back can be modelled as as ideal spring
(b) exerting a restoring force on the mass attached to its end.
(b) Still treating the bow string as if it were an ideal spring, we note that in pulling the
string from a displacement of x = 0 to x = 0.400 m the string does as amount of work on
given by Eq. 6.12:
= 0 -
= -46.0 J
Is this answer to the question? Not quite. . . we were really asked for the work done by the
hand on the bow string
. But at all times during the pulling, the hand exerted an equal and
opposite force on the string. The force had the opposite direction, so the work that it did
has the opposite sign. The work done (by the hand) in pulling the bow is +46.0 J.
The WorkKinetic Energy Theorem
7. A 40 kg box initially at rest is pushed 5.0 m along a rough horizontal floor with
a constant applied horizontal force of 130 N. If the coefficient of friction between
the box and floor is 0.30, find (a) the work done by the applied force, (b) the
energy lost due to friction, (c) the change in kinetic energy of the box, and (d)
the final speed of the box.
(a) The motion of the box and the forces which do work on it are shown in Fig. 6.3(a). The
(constant) applied force points in the same direction as the displacement. Our formula for
the work done by a constant force gives
= F d cos = (130 N)(5.0 m) cos 0
= 6.5 × 10
The applied force does 6.5 × 10
J of work.
(b) Fig. 6.3(b) shows all the forces acting on the box.
6.12: W string = 1 2 kx 2 i - 1 2 kx 2 f = 0 - 1 2 (575 N m )(0.400 m) 2 = -46.0 J Is this answer to the question? Our formula for the work done by a constant force gives W app = F d cos = (130 N)(5.0 m) cos 0 = 6.5 × 10 2 J The applied force does 6.5 × 10 2 J of work.