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If you push vigorously against a brick wall, how much work do you do on the wall? Explain. There is no work done on the wall as there is no displacement of the wall. Work = W = FsCosQ where: Work will be zero if F or S = 0 or angle Q = 90o In the current situation the wall is not moving and S is therefore zero. It follows that Work is zero, that is no work is done. A 0.2 kg rubber ball is dropped from the window of a building. It strikes the sidewalk below at 30 m/s and rebounds at 20 m/s. Find the change in momentum of the ball as a result of the collision with the side walk (in kg m/s). Mass of the rubber ball = 0.2 kg = m Initial velocity = 30 m/s = v1 Final velocity after collision = -20m/s = v2 Change in momentum = mv1 - mv2 = m(v1 - v2) = 0.2[30 - (-20)] = 10 kg m /s Q. A 2.0 kg block is thrown upward from a point 20m above Earth's surface. At what height above Earth's surface will the gravitational potential energy of the Earth block system have increased by 500J? Ans. Mass of the block m = 2.0 kg P.E. of the block at a height of 20 m is = mgh = 2 x 9,8 x 20 = 392 J Increase in P.E. = 500 J Let the new height be h New P.E. = mgh mgh -392 = 500 mgh = 500 + 392 = 892 J therefore h = 892/2x9.8 = 45.51m |
A person pushes a 10 kg cart a distance of 20 meters by exerting a 60 Newton horizontal force. The frictional resistance force is 50 Newtons. How much work is done by each force acting ont he cart? How much kinetic energy does the cart have at the end of the 20 meters if it started from rest:
Mass of cart = 10 kg
Distance moved = 20 m
Horizontal force = 60 N
Frictional force = 60 N
Frictional resistance force = 50 N
Net force = 60 - 50 = 10 N
Work done by the horizontal force = 60 x 20 = 1200 J
Work done by the frictional force = - 50 x 20 = -1000 J
(It is negative because force and displacement are opposite to each other)
Net work done = 1200 - 1000 = 200 J
= Change in the K.E. of the cart.
If the cart started from rest, the initial K.E. = 0
Final K.E. - Initial K.E. = 200 J
So Final K.E. = 200J
Assuming an efficiency of 25% for the muscle system in the process of converting food energy into mechanical work, how much energy would be used by a person of mass 75 kg (weight 165lb) in the process of climbing four flights of stairs for a total height of 15 metes? Find the answer in joules and convert to dietary calories (1 dietary calorie = 4186 Joules)
Efficiency ? = 25%
Mass of the person = 75 kg
Height climbed = h = 15 m
Work done = gain in P.E. = mgh
= 75 x 9.8 x 15 = 11025 J
Since the efficiency of the muscle system = 25%, the food energy required to do this work = (100 x 11025)/25 = 44100 J
4186 Joules = 1 dietary calorie
44100 Joules = 44100 / 4186 = 10.5 dietary calories.
The potential energy of a 0.2 kg particle moving along the x axis is given by:
U(x) = 8x2 - 2x4
where U is in Joules and x in meters. When the particle is at x = 10 m, what will be its acceleration?
m = 0.2 kg
P.E. U(x) = 8x2 - 2 x4
here U is in Joules and x in meters
Intensity T = -du/dx
=- -d[8x2 - 2 x4]/dx = -(16x -8x)
The intensity is equal to the force per unit mass which is the same as acceleration
a = -(16x - 8x)|x = 10 m
= -(160 -80) = -80 m/s2
In common speech, we often
use the words energy and power interchangeably as
if they were synonyms. However in physics the have
distinctly different definitions and are not synonymous.
Explain the difference between these terms as they are used in
physics.