MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
Physics 8.01
IC-W06D2-3 Work, Constant Forces and Scalar Product Solution
An object of mass m , starting from rest, slides down an inclined plane of length s . The
plane is inclined by an angle of  to the horizontal. The coefficient of kinetic friction
between the surface and the block is  k . What is the kinetic energy of the object after it
slides down the inclined plane a distance s ?
Solution While the object is sliding down the inclined plane the kinetic energy is
increasing due to the positive work done on the object by the gravitational force and the
negative work (smaller in magnitude) done by friction. The free body diagram is shown
below.
Choose a coordinate system with the origin at the top of the inclined plane and the
positive x -direction pointing down the inclined plane.
r
j and it is perpendicular to the displacement so it
The normal force is given by N  N φ
does no work,
r r
N  ds  N φ
j  dx φ
i0
(1.1)
The magnitude N of the normal force is determined from Newton’s Second Law applied
to the normal direction to the inclined plane,
N  mg cos  0 .
(1.2)
The friction force is then
r
f k   k N φ
i   k mg cos φ
i
(1.3)
The work done by the friction force is then
Wfriction
f
x f s
i
xi 0
r
r
  fk  d s 

(  k mg cos φ
i)  dx φ
i
.
xf s


(1.4)
 k mg cos dx    k mg cos s  0
xi 0
The gravitational force is given by
r
Fgrav  mg sin  φ
i  mg cos φ
j
(1.5)
The work done by the gravitational force is
Wfriction
f
xf s
i
xi 0
r
r
  Fgrav  d s 

(mg sin  φ
i  mg cos φ
j)  dx φ
i
(1.6)
xf s


mg sin  dx  mg sin  s  0
xi 0
The sum of the work is
W  Wgrav  Wfriction  mgs(sin  k cos )
(1.7)
which is positive because the object is increasing its kinetic energy.
The object started from rest, and so the change in kinetic energy is equal to the final
kinetic energy at the bottom of the incline,
K 
1
mv f 2
2
(1.8)
The work-kinetic energy theorem
W  K
(1.9)
thus substituting Eqs. (1.7) and (1.8) into Eq. (1.9) yields
mgs(sin   k cos ) 
1 2
mv
2 f
(1.10)