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Dafna1 [17]
3 years ago
13

If it requires 8.0 J of work to stretch a particular spring by 2.0 cm from its equilibrium length, how much more work will be re

quired to stretch it an additional 4.0 cm? Express your answer to two significant figures and include the appropriate units.
Physics
1 answer:
jenyasd209 [6]3 years ago
7 0

Answer:

The amount of work done required to stretch spring by additional 4 cm is 64 J.

Explanation:

The energy used for stretching spring is given by the relation :

E = \frac{1}{2}kx^{2}           .......(1)

Here k is spring constant and x is the displacement of spring from its equilibrium position.

For stretch spring by 2.0 cm or 0.02 m, we need 8.0 J of energy. Hence, substitute the suitable values in equation (1).

8 = \frac{1}{2}\timesk\times k \times(0.02)^{2}

k = 4 x 10⁴ N/m

Energy needed to stretch a spring by 6.0 cm can be determine by the equation (1).

Substitute 0.06 m for x and 4 x 10⁴ N/m for k in equation (1).

E = \frac{1}{2}\times4\times10^{4}\times (0.06)^{2}

E = 72 J

But we already have 8.0 J. So, the extra energy needed to stretch spring by additional 4 cm is :

E = ( 72 - 8 ) J = 64 J

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A spherical, conducting shell of inner radius r1= 10 cm and outer radius r2 = 15 cm carries a total charge Q = 15 μC . What is t
lutik1710 [3]

a) E = 0

b) 3.38\cdot 10^6 N/C

Explanation:

a)

We can solve this problem using Gauss theorem: the electric flux through a Gaussian surface of radius r must be equal to the charge contained by the sphere divided by the vacuum permittivity:

\int EdS=\frac{q}{\epsilon_0}

where

E is the electric field

q is the charge contained by the Gaussian surface

\epsilon_0 is the vacuum permittivity

Here we want to find the electric field at a distance of

r = 12 cm = 0.12 m

Here we are between the inner radius and the outer radius of the shell:

r_1 = 10 cm\\r_2 = 15 cm

However, we notice that the shell is conducting: this means that the charge inside the conductor will distribute over its outer surface.

This means that a Gaussian surface of radius r = 12 cm, which is smaller than the outer radius of the shell, will contain zero net charge:

q = 0

Therefore, the magnitude of the electric field is also zero:

E = 0

b)

Here we want to find the magnitude of the electric field at a distance of

r = 20 cm = 0.20 m

from the centre of the shell.

Outside the outer surface of the shell, the electric field is equivalent to that produced by a single-point charge of same magnitude Q concentrated at the centre of the shell.

Therefore, it is given by:

E=\frac{Q}{4\pi \epsilon_0 r^2}

where in this problem:

Q=15 \mu C = 15\cdot 10^{-6} C is the charge on the shell

r=20 cm = 0.20 m is the distance from the centre of the shell

Substituting, we find:

E=\frac{15\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.20)^2}=3.38\cdot 10^6 N/C

4 0
3 years ago
A 0.02kg dart is loaded into a toy spring gun by compressing the spring 6cm. If the spring has a constant of 20 N/m, find… a. Th
katrin2010 [14]

Answer:

a) 0.036 J b) 0.036J c) 0.036 d) 1.9m/s e) 0.18 m

Explanation:

Mass of the dart = 0.02kg, the spring was compressed to 6cm

Work needed to compress the spring = 1/2*k*x ^2 where k is the force constant of the spring in N/m, x is the distance it was compressed in m

Work needed to compress the spring = 0.5 * 20* 0.06^2 since 6cm = 6 / 100 = 0.06 m

Work needed to compress the spring = 0.036J

b) the total energy stored in the spring = the work done to compress the spring = 0.036J

c) kinetic energy of the dart as it leaves the the spring = elastic potential energy stored in the spring = the work done in compressing the = 0.036J using the law of conservation of energy; energy is neither created nor destroyed but transformed from one form to another.

d) 1/2mv^2 = 0.036

mv^2 = 0.036*2

v^2 = 0.036*2 / 0.02 = 3.6

v = √3.6 = 1.897 approx 1.9m/s

e) kinetic energy of the dart = work done against gravity to get the body to height h

Work done against gravity = potential energy conserved at height = -mgh g is negative because the motion is upward while gravity acts downward

0.036 = 0.02 * 9.81 * h

0.036 / ( 0.02*9.81) = h

h = 0.18 m

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