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These results are exactly the same as for the vertical motion discussed in the previous section. Interestingly, work along any incline that reaches same height is same. The objects released from a given height will have same velocity and kinetic energy, irrespective of the length of incline involved.

Motion along incline

Blocks have same velocity and kinetic energy on reaching ground.

The work, kinetic energy and velocity are independent of path between end points, provided the vertical height is same in all cases. This is even true for curved path. Any curved path can be considered to be series of linear path having horizontal and vertical components. As horizontal component is not relevant, vertical components sum up to the net height.

Motion along curved path

Vertical components sum up the net height.

We must, however, emphasize that this is not a general result, but specific to gravity and other conservative forces (we shall define this concept in a separate module.). For example, if the surfaces are rough, then block will not be able to retain its kinetic energy on the return to initial position of projection due to friction, which is not a conservative force.

In the nutshell, we have learned following characterizing features of the motion of block acted upon by gravity alone :

  • Gravity is a constant force. Hence, work by gravity is constant for a given displacement.
  • The particle retains the kinetic energy on return to same vertical position.
  • Work by gravity for horizontal displacement is zero as force and displacement are at right angle.
  • Work by gravity is independent of path details and depends only on the vertical component of displacement.

Particle is subjected to gravity and other forces

We must understand that work by force due to gravity for a given displacement is independent of the presence of other forces. The work done by gravity remains same. When other forces are also present, “work-kinetic energy” theorem has following form :

K f - K i = W G + W F

where W G and W F are the work done by the gravity and other force(s).

Here, work done by other external force(s) may be analyzed with respect to following different conditions :

  • Initial and final speeds are zero.
  • Initial and final speeds are same.
  • Initial and final speeds are different.

In the first two cases, initial and final kinetic energy are same. Hence,

K f - K i = W G + W F = 0 W F = - W G

This is an important result. The work done by other force(s) is same as that by gravity, but with a negative sign (we have already discussed such case in previous module). This feature of motion, when end conditions are same, facilitates computation of work by other forces a great deal, as we need not be concerned of the external force other than gravity. We simply compute the work by gravitational force. The work by other external force(s) is equal to the work by gravity with a negative sign.

In third case, kinetic energies at end points are not same. The difference in kinetic energy during a motion is equal to the net work done by all forces. In this case, we can not consider work by other forces to be equal in magnitude to that by gravity. Instead, we would require to find individual force by force analysis and then compute work by each force individually.

Problem : A lift, weighing 1000 kg, is raised vertically by a string - pulley arrangement with an upward acceleration of 1 m / s 2 . If the initial speed of the lift is 2 m/s, find the speed attained by the lift after traveling a vertical distance of 10 m.

Solution : Here, lift is moving with upward acceleration of 1 m / s 2 . It means that velocities in the beginning and end are not same. Therefore, work done by gravity and tension is not zero.

Now, work done by gravity is :

W G = - m g y = - 1000 x 10 x 10 = - 100000 J

In order to know the work done by the tension, we need to know the tension. Note here that work by tension is not equal to work by gravity as in earlier case. From, free body diagram, we have :

Free body diagram

T - m g = m a T = m ( g + a ) = 1000 x ( 10 + 1 ) = 11000 N

Work done by tension in the string is :

W T = T x y W T = 11000 x 10 = 110000 J

Using “work - kinetic energy” theorem, we have :

K f - K i = W G + W F = - 100000 + 110000 = 10000 J 1 2 m ( v f 2 - v i 2 ) = 10000

v f 2 = 2 x 10000 m + v i 2

v f 2 = 20 + 2 2 v f 2 = 24 v f = 4.9 m / s

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Questions & Answers

How can we take advantage of our knowledge about motion?
Kenneth Reply
pls explain what is dimension of 1in length and -1 in time ,what's is there difference between them
Mercy Reply
what are scalars
Abdhool Reply
show that 1w= 10^7ergs^-1
Lawrence Reply
what's lamin's theorems and it's mathematics representative
Yusuf Reply
if the wavelength is double,what is the frequency of the wave
Ekanem Reply
What are the system of units
Jonah Reply
A stone propelled from a catapult with a speed of 50ms-1 attains a height of 100m. Calculate the time of flight, calculate the angle of projection, calculate the range attained
Samson Reply
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water boil at 100 and why
isaac Reply
what is upper limit of speed
Riya Reply
what temperature is 0 k
0k is the lower limit of the themordynamic scale which is equalt to -273 In celcius scale
How MKS system is the subset of SI system?
Clash Reply
which colour has the shortest wavelength in the white light spectrum
Mustapha Reply
how do we add
Jennifer Reply
if x=a-b, a=5.8cm b=3.22 cm find percentage error in x
Abhyanshu Reply
x=5.8-3.22 x=2.58
what is the definition of resolution of forces
Atinuke Reply

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