The weight of a block on the inclined plane is 500 N and the angle of incline is 30 degrees. What is the magnitude of the force
of friction for static equilibrium? What is the normal force of the surface pushing up on the block?
1 answer:
Answer:
250 N
433 N
Explanation:
N = Normal force by the surface of the inclined plane
W = Weight of the block = 500 N
f = static frictional force acting on the block
Parallel to incline, force equation is given as
f = W Sin30
f = (500) Sin30
f = 250 N
Perpendicular to incline force equation is given
N = W Cos30
N = (500) Cos30
N = 433 N
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<h2>Answer: 1000 J</h2>
The Work done by a Force
refers to the release of potential energy from a body that is moved by the application of that force to overcome a resistance along a path.
It should be noted that it is a scalar magnitude, and its unit in the International System of Units is the Joule (like energy). Therefore, 1 Joule is the work done by a force of 1 Newton when moving an object, in the direction of the force, along 1 meter:
Now, when the applied force is constant and the direction of the force and the direction of the movement are parallel, the equation to calculate it is:
(1)
When they are not parallel, both directions form an angle, let's call it . In that case the expression to calculate the Work is:
(2)
For example, in order to push the 200 N box across the floor, you have to apply a force along the distance to overcome the resistance of the weight of the box (its 200 N).
In this case both <u>(the force and the distance in the path) are parallel</u>, so the work performed is the product of the force exerted to push the box
by the distance traveled
. as shown in equation (1).
Hence:
>>>>This is the work