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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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.