The force that prevents motion when the surfaces of two objects come into contact is known as friction. Friction decreases a machine's mechanical advantage, or, to put it another way, reduces the output to input ratio.
<h3>How can I figure out the frictional force?</h3>
The resistive force of friction (Fr) divided by the normal or perpendicular force (N) pushing the objects together yields the coefficient of friction (fr), which is a numerical value.
The formula fr = Fr/N serves as a representation of it.
Therefore, 100N of force is needed to move an item with a mass of 50 kg.
It will accelerate by 10 m/s2.
If a substance's mass does not change over time, friction cannot affect it. Instead, friction can be affected in a variety of ways by an object's mass.
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Air resistance force
tension force
spring force
frictional force
normal force
gravitational force
applied force
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(a) The magnitude and direction of the net force on the crate while it is on the rough surface is 36.46 N, opposite as the motion of the crate.
(b) The net work done on the crate while it is on the rough surface is 23.7 J.
(c) The speed of the crate when it reaches the end of the rough surface is 0.45 m/s.
<h3>Magnitude of net force on the crate</h3>
F(net) = F - μFf
F(net) = 280 - 0.351(92 x 9.8)
F(net) = -36.46 N
<h3>Net work done on the crate</h3>
W = F(net) x L
W = -36.46 x 0.65
W = - 23.7 J
<h3>Acceleration of the crate</h3>
a = F(net)/m
a = -36.46/92
a = - 0.396 m/s²
<h3>Speed of the crate</h3>
v² = u² + 2as
v² = 0.845² + 2(-0.396)(0.65)
v² = 0.199
v = √0.199
v = 0.45 m/s
Thus, the magnitude and direction of the net force on the crate while it is on the rough surface is 36.46 N, opposite as the motion of the crate.
The net work done on the crate while it is on the rough surface is 23.7 J.
The speed of the crate when it reaches the end of the rough surface is 0.45 m/s.
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Answer:
C. The spring constant of spring 2 is larger than spring 1.
Explanation:
As we know that spring constant also called the stiffness constant of a spring is the value of force required for the deformation of the spring by a unit length.
Mathematically given as:
where:
stiffness constant
force on the spring
change in length of the spring
It is given in the question that spring 1 is easily compressible and requires less force to compress as compared to spring 2 and hence it has lesser stiffness constant than that of spring 2.
Answer:
Electric motors build energy through the interaction between magnetic fields.
Magnetic fields may be generated by passing a current through a wire. Winding that wire into a coil intensifies the sector. Wrapping the coil around an iron core can intensify it more. One is going to be of the coil are North, the other South. Reverse the flow of current through the coil. The magnetic poles can swap. Place 2 separate coils close to one another and organize them therefore one turns whereas the opposite is fastened. Then build arrangements for this within the moving coil to reverse even as the opposite poles are about to line up.
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