Answer:
Work done on an object is equal to
FDcos(angle).
So, naturally, if you lift a book from the floor on top of the table you do work on it since you are applying a force through a distance.
However, I often see the example of carrying a book through a horizontal distance is not work. The reasoning given is this: The force you apply is in the vertical distance, countering gravity and thus not in the direction of motion.
But surely you must be applying a force (and thus work) in the horizontal direction as the book would stop due to air friction if not for your fingers?
Is applying a force through a distance only work if causes an acceleration? That wouldn't make sense in my mind. If you are dragging a sled through snow, you are still doing work on it, since the force is in the direction of motion. This goes even if velocity is constant due to friction.
Explanation:
The gravitational force on the woman is A) 500 N
Explanation:
There are two forces acting on the woman during her fall:
- The force of gravity,
, acting downward - The air resistance,
, acting upward
According to Newton's second law, the net force acting on the woman is equal to the product between the woman's mass and her acceleration:
where m is the mass of the woman and a her acceleration.
The net force can be written as
Also, we know that the woman falls at a constant velocity (5 m/s), this means that her acceleration is zero:
Combining the equations together, we get:
which means that the magnitude of the gravitational force is equal to the magnitude of the air resistance:
Learn more about forces and Newton's second law:
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Answer:
The hiker followed a road heading north for 2 miles in 30 minutes.
Explanation:
In order to describe the motion of an object, distance covered and time taken must be required. The total path covered by an object is called the distance travelled.
The hiker followed a road heading north for 2 miles in 30 minutes. This describes the motion of hiker. The motion shows how fast the hiker is moving.
Distance, d = 2 miles = 3218.6 m
times, t = 30 minutes = 1800 seconds
So, we can say that the hiker is moving with a speed of 1.78 m/s in north direction.
Hence, this is the required solution.
1.5 m/s is the velocity.
9.3 m is the length of aisle, over which Distance will be covered.
Time is demanded in which the child will move the cart over the aisle with 1.5 m/s.
v=S/t
and,
t=S/v
Put values,
t=9.3/1.5=6.2 s
