Aceiling fan is turned on and reaches an angular speed of 120 rev/min in 20 s. it is then turned off and coasts to a stop in an additional 60 s. the ratio of the average angular acceleration for the first 20 s to that for the last 60 s is which of the following?
What is the kinetic energy of a baseball with a mass 2 kg moving at a speed of 4 m/s?
m (mass) = 2 kg
v (velocity) = 4 m/s
K.E. (kinetic energy) = ? unknown
The formula for Kinetic Energy is, K.E. = 1/2mv²
K.E. = 1/2mv²
= 1/2 × 2 kg × (4 m/s)²
= 1/2 × 2 kg × 16 m²/s²
= 16 kg.m²/s²
= 16 J
16 J or 16 kg.m²/s²
In this physics problem, you will be dealing with work. Work is the scalar product or dot product of two vector quantities. These quantities are force (F) and displacement (d). Any quantity resulting from the dot product of two vectors is a scalar quantity. Therefore, it should be clear at this point that work is a scalar quantity. Also, the unit for work is "Joule" (J) which is equivalent to the unit "Newton-meter" (N·m).
In equation form, work is expressed as
W = F · d
This can also be represented as
W = (Fcos∅)d
where Fcos∅ is the component of the constant force applied to the object in the direction of the displacement.
Question: How much work does each of the following forces on the book?
A. Your 2.40 N push (F)
Use the equation
W = (Fcos∅)d where ∅ = 0° (This is because the direction of displacement and direction of force are the same, that is, along the horizontal axis.)
Recall that cos(0°) = 1. So the equation will become
W = Fd
The magnitude of force (F) is 2.40 N while the magnitude of displacement (d) is 1.50 m. Substituting these values in the equation, you will have
W = (2.40 N)(1.50 m)
W = 3.6 N·m or 3.6 J
Therefore, the work done by your 2.40 N push on the book is 3.60 J.
B. Friction force (f)
You will use the equation
W = fdcos∅ where ∅ = 180° because the direction of friction is opposite to the direction of displacement.
Recall that cos(180°) = -1. Hence, the equation will become
W = -fd
It is given that the magnitude of friction (f) is 0.600 N while the magnitude of displacement is still 1.50 m. Substituting these values, we'll get
W = (-0.600 N)(1.50 m)
W = -0.900 N·m 0r -0.900 J
Therefore, the work done by friction to the book is -0.900 J.
C. Normal force
Any force that is perpendicular to the direction of displacement exerts no work on the object. Since the normal force is perpendicular to the direction of displacement, therefore work is zero.
W = 0 J
Any force that is perpendicular to the direction of displacement exert no work on the object. Since gravity is perpendicular to the direction of displacement, therefore work is zero.
W = 0 J
For examples of situations involving work, you can check Sometimes, you will encounter problems where force is expressed in vector form and you still have to compute for its magnitude. An example of this is provided in A very simple and straightforward example of work problem is also given in .
Legend: w=Work f=Force d=Distance
w=200n * 20m
4.)Substitute nm to Joules