If a load of 100N is applied to a wire,the extension is 0.01m. Calculate the energy stored in the wire.
do
this the same way as the one below this
k = F/x
and
energy stored = work done = (1/2) k x^2
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Now I know but I think everyone is equal in brain so I will do not going to give answers of your questions
If a load of 1000N is applied to a wire the extension is 0.01m calculate the energy stored in the wire.
We can use the equation for the spring constant:
k = F/x = 1000 N / 0.01 m = 100,000 N/m
And the equation for the energy stored in a spring:
E = (1/2) k x^2 = (1/2) * 100,000 N/m * (0.01 m)^2 = 5 J
Therefore, the energy stored in the wire is 5 joules.
A spring of national length 3m is extended by 0.01m by a force of 4N what will be the length when the applied force is2N
We can use the equation for the spring constant:
k = F/x = 4 N / 0.01 m = 400 N/m
Using Hooke's law, we can set up the equation:
F = k * (L - L0)
Where F is the force, k is the spring constant, L is the length of the spring when the force is applied, and L0 is the natural length of the spring.
When the force is 4N:
4 N = 400 N/m * (L - 3 m)
L - 3 m = 0.01 m
L = 3.01 m
So, the length of the spring when the force is 4N is 3.01m.
When the force is 2N:
2 N = 400 N/m * (L - 3 m)
L - 3 m = 0.005 m
L = 3.005 m
So, the length of the spring when the force is 2N is 3.005m.
A uniform steel wire has crossed sectional area of 10×10^-6m and length of 10m.A Mass of 10kg is attached at one end. If the young modulus of steel is 10^-2nm^-2 Calculate the extension produced takes gravity at 9.8m/s^-2
We can calculate the elongation of the steel wire using the formula:
stress = (force) / (cross-sectional area)
strain = (extension) / (original length)
Young's modulus is given by:
Young's modulus = (stress) / (strain)
Combining these equations, we can write:
(extension) = (force * length) / (Young's modulus * cross-sectional area)
The weight of the mass is given by:
(force) = (mass) * (acceleration due to gravity)
Substituting the values given:
(force) = (10 kg) * (9.8 m/s^2) = 98 N
(extension) = (98 N * 10 m) / (10^-2 N/m^2 * 10 × 10^-6 m^2) = 9.8 × 10^-4 m = 0.98 mm
Therefore, the extension produced is 0.98 mm.
A body of Mass 2kg is attached to the end of a vertical wire of length 2m diameter 0.64mm and the extension of 0.6 calculate the tensile stress and tensile strain.
The cross-sectional area of the wire can be calculated using the formula:
area = (π/4) x diameter^2
area = (π/4) x (0.64 mm)^2 = 0.32256 × 10^-6 m^2
The weight of the body is given by:
force = mass x acceleration due to gravity
force = 2 kg × 9.8 m/s^2 = 19.6 N
The tensile stress can be calculated using the formula:
stress = force / area
stress = 19.6 N / 0.32256 × 10^-6 m^2 = 6.06 × 10^7 Pa
The tensile strain can be calculated using the formula:
strain = extension / original length
strain = 0.6 m / 2 m = 0.3
Therefore, the tensile stress is 6.06 × 10^7 Pa and the tensile strain is 0.3.
A load of 15N is placed on a scale attached spring balance.this load causes an extension of 5cm in the spring. What load must be placed on the spring balance to cause an extension of 8cm.is the spring constant of spring?
We can use the formula for the spring constant, which relates the force applied to the spring and its resulting extension:
k = F / x
where k is the spring constant, F is the force applied, and x is the extension. We know that the spring constant is constant, so we can use it to find the force required for a different extension by rearranging the formula:
F = k x
First, we can calculate the spring constant from the first scenario:
k = F / x = 15 N / 0.05 m = 300 N/m
Now we can use this spring constant to find the force required for the new extension:
F = k x = 300 N/m × 0.08 m = 24 N
Therefore, a load of 24 N must be placed on the spring balance to cause an extension of 8 cm.
(Note: The units for extension in the first scenario were given in centimeters, but it's important to convert to meters for use in the formula for the spring constant.)
A spiral loaded with a price of metal extended by10.5cm in air when the metal is fully submerged in water the spring extends by 6.8cm.Calculate the relative density of the metal assuming Hoke's law is obeyed.
We can use Hooke's law, which states that the extension of a spring is proportional to the force applied, to find the relative density of the metal. The force on the spring is due to the weight of the metal, and we know that the weight of an object is equal to its mass times the acceleration due to gravity. Therefore:
force in air = (mass of metal) x (acceleration due to gravity)
force in water = (mass of metal - mass of displaced water) x (acceleration due to gravity)
Using Hooke's law, we can set up the equation:
k x = (force in air) = (force in water)
where k is the spring constant and x is the difference in extension between the two scenarios.
Therefore:
k x = (mass of metal) x (acceleration due to gravity)
k (x + 3.7 cm) = (mass of metal - mass of displaced water) x (acceleration due to gravity)
where 3.7 cm is the difference in extension due to the buoyant force of the water.
Dividing the second equation by the first equation, we get:
x + 3.7 cm = (mass of metal - mass of displaced water) / (mass of metal) x x
Solving for the ratio of mass to displaced water, we get:
(mass of metal - mass of displaced water) / (mass of metal) = (x + 3.7 cm) / x
(mass of metal - mass of displaced water) / mass of metal = 0.981 (since x = 10.5 cm and x + 3.7 cm = 14.2 cm)
The density of water is 1000 kg/m^3. Since the buoyant force is equal to the weight of the displaced water, we can calculate the mass of the displaced water:
mass of displaced water = density of water x volume of displaced water
mass of displaced water = 1000 kg/m^3 x (x + 3.7 cm) x (10^-2 m/cm)^3
The volume of the metal is the same as the volume of the displaced water, so we can calculate the mass of the metal as:
mass of metal = density of metal x volume of metal
mass of metal = (density of water x x) x (10^-2 m/cm)^3
Substituting these values into the equation for the ratio of mass to displaced water, we get:
(density of water x x) x (10^-2 m/cm)^3 - (density of water x x) x (10^-2 m/cm)^3 / (density of water x x) x (10^-2 m/cm)^3 = 0.981
Simplifying, we get:
density of water / density of metal = 0.981
Therefore, the relative density of the metal is approximately 1.019.
A wire is gradually stretched unit it's shape sketch a load of extension graph for the wire and the on the graph indicates elastic limit, the yield point the maximum load and the breaking point.
Here is a sketch of a typical load-extension graph for a wire:
Load
| ^
| |
| | Plastic Deformation
| | (Beyond Yield Point)
| | |
| | Elastic | Necking
| | Deform. | |
____| |______________ | V
| | |<- Elastic|---|-------|---|--------->
| | | Deformation | Yield | Breaking
| | |____________|
| | Slope: k = F/x
| |
| |---> Extension
| |
| |
| -
| \
| \
| \
| \
| \
| \
| \
| \
| \
|
0 Extension
The graph shows the relationship between the load (force applied) and the extension (change in length) of the wire. Initially, as the load is increased, the wire undergoes elastic deformation, meaning that it will return to its original shape when the load is removed. This is shown by the linear portion of the graph, which has a slope of k = F/x, where k is the spring constant of the wire.
As the load continues to increase beyond a certain point (the elastic limit), the wire will begin to undergo plastic deformation, meaning that it will undergo permanent changes to its shape. This is shown by the non-linear portion of the graph, which has a decreased slope due to the reduction in rigidity caused by the plastic deformation.
The yield point is the point on the graph where the wire begins to undergo significant plastic deformation. This is the point where the slope of the graph decreases from the elastic slope. Beyond the yield point, the wire is considered to have undergone plastic deformation.
The maximum load is the point on the graph where the wire can no longer support the load and begins to break. This is usually represented by a sudden drop in the load as the wire breaks.
The breaking point is reached when the wire has completely fractured and can no longer support any load.