Question 1

A mass on a spring vibrates in simple harmonic motion at a frequency of 4.0 Hz and an amplitude of 4.0 cm.If a timer is started when its displacement is a maximum (hence x = 4 cm when t = 0),what is the speed of the mass when t = 5 s?

Accepted Answer

The speed of the mass is zero when the displacement is at a maximum, which occurs at t = 5 s since the period is 0.25 s and 5 s is an integer multiple of the period.

Question 2

A car with bad shocks bounces up and down with a period of 1.50 s after hitting a bump.The car has a mass of 1250 kg and is supported by four springs of force constant k.What is k for each spring?

Accepted Answer

The period of oscillation of a spring-mass system is given by:

T = 2π√(m/k)

where T is the period, m is the mass, and k is the spring constant.

Rearranging the formula, we get:

k = (4π²m)/T²

Substituting given values, we have:

k = (4π² x 1250)/1.50²

k = 5479.5 N/m

which is closest to option B: 5480 N/m. Therefore, B is the best choice.

Question 3

Suppose a 0.30-kg mass on a spring-loaded gun that has been compressed 0.20 m has elastic potential energy of 1.0 J.How high above the spring's equilibrium point can the gun fire the mass if the gun is fired straight up?

Accepted Answer

The answer of Suppose a 0.30-kg mass on a spring-loaded...

Question 4

A 2.0 cm 
2)0 cm steel rod of length 80 cm has its left end clamped and a force of 
Applied perpendicular to its length at its right end.By how much does the right end shift from its original position? For steel Young's modulus is 
,the bulk modulus is 
,and the shear modulus is 
)

Accepted Answer

The right end of the steel rod will experience a displacement given by the formula:
$$\Delta L = \frac{F\cdot L}{A\cdot E}$$
where $F$ is the applied force, $L$ is the length of the rod, $A$ is the cross-sectional area of the rod, and $E$ is the Young's modulus of the steel. 
Substituting the given values, we have:
$$\Delta L = \frac{(40\ 	ext{N})\cdot (80\ 	ext{cm})}{(\pi\cdot (1\ 	ext{cm})^2)\cdot (2	imes 10^{11}\ 	ext{N/m}^2)} \approx 0.101\ 	ext{mm}$$
Therefore, the right end of the rod shifts by about 0.101 mm. 
The best choice is B, which matches this result.

Question 5

A uniform pressure of 21 × 10⁵ N/m² is applied to all six sides of a copper cube.What is the percentage change in volume of the cube? (for copper,B = 14 × 10¹⁰ N/m²)

Accepted Answer

The formula for percentage change in volume is ΔV/V = 3ΔL/L, where ΔL/L is the fractional change in length.

Since pressure (P) = stress (σ), we can use σ = Bε, where B is the bulk modulus and ε is the strain.

For a cube, we have ε = ΔL/L = −P/B. Substituting in the values,

ε = −(21 × 10^15 N/m^2)/(14 × 10^10 N/m^2) = −1.5 × 10^−5

ΔV/V = 3ε = −4.5 × 10^−5

The negative sign indicates a decrease in volume. Therefore, the percentage change in volume is:

ΔV/V × 100% = −1.5 × 10^−3%, which is approximately 1.5 × 10^−13%, or 1.5 × 10^−14 (rounded to one decimal place).

The closest option is B: 1.5 × 10^−13%.

Question 6

A copper alloy wire of length 2.0 m,cross sectional area 7.1 × 10−⁶ m²,and Young's modulus 11 × 10¹⁰ N/m²,has a 200-kg load hung on it.What is its increase in length? (g = 9.8 m/s²)

Accepted Answer

The answer of A copper alloy wire of length 2.0...

Question 7

How large a force is necessary to stretch a 2.0 mm radius steel wire (Y = 2.0 × 10¹¹ N/m²)by 2.0%?

Accepted Answer

The stress is:$$\sigma = Y \epsilon = (2.0 	imes 10^{11} 	ext{ N/m}^2)(2.0 	imes 10^{-2}) = 4.0 	imes 10^9 	ext{ N/m}^2$$The force is:$$F = \sigma A = (4.0 	imes 10^9 	ext{ N/m}^2)(\pi (2.0 	imes 10^{-3} 	ext{ m})^2) = 5.0 	imes 10^4 	ext{ N}$$

Question 8

An ore car of mass 2000 kg rolls downhill on tracks from a mine.At the end of the tracks,10.0 m lower in elevation,is a spring with k = 400,000 N/m.How much is the spring compressed in stopping the ore car? Ignore friction.

Accepted Answer

The answer of An ore car of mass 2000 kg...

Question 9

Suppose a 0.30-kg mass on a spring that has been compressed 0.20 m has elastic potential energy of 1.0 J.How much further must the spring be compressed to triple the elastic potential energy?

Accepted Answer

The elastic potential energy stored in the spring can be given by:

PE = 0.5*k*x^2

where PE is the elastic potential energy, k is the spring constant, and x is the displacement of the spring from its equilibrium position.

Given that the mass on the spring has an elastic potential energy of 1.0 J when it is compressed by 0.2 m, we can write:

1.0 J = 0.5*k*(0.2 m)^2

Solving for k, we get:

k = 25 N/m

Now, we need to find how much further the spring must be compressed to triple the elastic potential energy. Let's assume that the spring is compressed by x m from its equilibrium position to achieve this.

The new elastic potential energy stored in the spring can be given by:

PE' = 0.5*k*(x + 0.2 m)^2

We want PE' to be three times the original PE. Therefore, we can write:

3*1.0 J = 0.5*(25 N/m)*(x + 0.2 m)^2

Simplifying and solving for x, we get:

x = 0.15 m

Therefore, the spring must be compressed by an additional 0.15 m to triple the elastic potential energy stored in it. The best choice is B) 0.15 m.

Question 10

To double the total energy of a simple harmonic oscillator,by what factor should the amplitude change?

Accepted Answer

The answer of To double the total energy of a...

Question 11

The oxygen molecule (O₂)may be regarded as two masses connected by a spring.In vibrational motion,each oxygen atom alternately approaches,then moves away from the center of mass of the system.If each oxygen atom of mass m = 2.67 × 10−²⁶ kg has a vibrational energy of 6.4 × 10−²¹ J and the effective spring constant is 50 N/m,then what is the amplitude of oscillation of each oxygen atom?

Accepted Answer

The vibrational energy of the oxygen atom is given by:

E = 1/2 kA^2

where k is the effective spring constant and A is the amplitude of oscillation.

Rearranging the equation, we get:

A = sqrt(2E / k)

Substituting the values given in the problem, we get:

A = sqrt((2 x 6.4 x 10^-21 J) / 50 N/m)
  = 1.6 x 10^-11 m

Since each oxygen atom oscillates around the center of mass, the amplitude of oscillation for each oxygen atom is half of this value:

Amplitude of oscillation for each oxygen atom = 0.8 x 10^-11 m = 1.6 x 10^-11 / 2 = 8.0 x 10^-12 m

Therefore, the best choice is B: 1.6 x 10^-11 m.

Question 12

A 0.20-kg block rests on a frictionless,level surface and is attached to a horizontally aligned spring with a spring constant of 80 N/m.The block is initially displaced 4.0 cm from the equilibrium point and then released to set up a simple harmonic motion.A frictional force of 0.3 N exists between the block and surface.What is the speed of the block when it passes through the equilibrium point after being released from the 4.0-cm displacement point?

Accepted Answer

Using conservation of mechanical energy, we can find the maximum speed of the block at the equilibrium point. The potential energy stored in the spring when the block is displaced 4.0 cm is (1/2)(k)(x^2) = (1/2)(80)(0.04) = 1.6 J. At the equilibrium point, all of the potential energy is converted to kinetic energy, so (1/2)(m)(v^2) = 1.6 J. Solving for v, we get v = sqrt((2*1.6)/0.20) = 1.8 m/s. However, we must account for the frictional force which acts in the opposite direction of motion. The work done by friction is F*d = (0.3)(0.04) = 0.012 J. This reduces the amount of mechanical energy available, so the kinetic energy at the equilibrium point is (1.6 - 0.012) J. Solving for v again gives v = sqrt((2*1.588)/0.20) = 0.72 m/s, which is answer choice C.

Question 13

I attach a 2.0-kg block to a spring that obeys Hooke's Law and supply 4.0 J of energy to stretch the spring.I release the block;it oscillates with period 0.30 s.The amplitude is:

Accepted Answer

The answer of I attach a 2.0-kg block to a...

Question 14

Suppose a 0.30-kg mass on a spring that has been compressed 0.10 m has elastic potential energy of 1.5 J.What is the spring constant?

Accepted Answer

The answer of Suppose a 0.30-kg mass on a spring...

Quiz 11: Energy in Thermal Processes Heat and Internal Energy: Part A

A mass on a spring vibrates in simple harmonic motion at a frequency of 4.0 Hz and an amplitude of 4.0 cm.If a timer is started when its displacement is a maximum (hence x = 4 cm when t = 0) ,what is the speed of the mass when t = 5 s?

A car with bad shocks bounces up and down with a period of 1.50 s after hitting a bump.The car has a mass of 1250 kg and is supported by four springs of force constant k.What is k for each spring?

Suppose a 0.30-kg mass on a spring-loaded gun that has been compressed 0.20 m has elastic potential energy of 1.0 J.How high above the spring's equilibrium point can the gun fire the mass if the gun is fired straight up?

A 2.0 cm 2) 0 cm steel rod of length 80 cm has its left end clamped and a force of Applied perpendicular to its length at its right end.By how much does the right end shift from its original position? For steel Young's modulus is ,the bulk modulus is ,and the shear modulus is )

A uniform pressure of 21 × 10⁵ N/m² is applied to all six sides of a copper cube.What is the percentage change in volume of the cube? (for copper,B = 14 × 10¹⁰ N/m²)

A copper alloy wire of length 2.0 m,cross sectional area 7.1 × 10−⁶ m²,and Young's modulus 11 × 10¹⁰ N/m²,has a 200-kg load hung on it.What is its increase in length? (g = 9.8 m/s²)

How large a force is necessary to stretch a 2.0 mm radius steel wire (Y = 2.0 × 10¹¹ N/m²) by 2.0%?

An ore car of mass 2000 kg rolls downhill on tracks from a mine.At the end of the tracks,10.0 m lower in elevation,is a spring with k = 400,000 N/m.How much is the spring compressed in stopping the ore car? Ignore friction.

Suppose a 0.30-kg mass on a spring that has been compressed 0.20 m has elastic potential energy of 1.0 J.How much further must the spring be compressed to triple the elastic potential energy?

To double the total energy of a simple harmonic oscillator,by what factor should the amplitude change?

I attach a 2.0-kg block to a spring that obeys Hooke's Law and supply 4.0 J of energy to stretch the spring.I release the block;it oscillates with period 0.30 s.The amplitude is:

Suppose a 0.30-kg mass on a spring that has been compressed 0.10 m has elastic potential energy of 1.5 J.What is the spring constant?

Quiz 11: Energy in Thermal Processes Heat and Internal Energy: Part A

A mass on a spring vibrates in simple harmonic motion at a frequency of 4.0 Hz and an amplitude of 4.0 cm.If a timer is started when its displacement is a maximum (hence x = 4 cm when t = 0) ,what is the speed of the mass when t = 5 s?

A car with bad shocks bounces up and down with a period of 1.50 s after hitting a bump.The car has a mass of 1250 kg and is supported by four springs of force constant k.What is k for each spring?

Suppose a 0.30-kg mass on a spring-loaded gun that has been compressed 0.20 m has elastic potential energy of 1.0 J.How high above the spring's equilibrium point can the gun fire the mass if the gun is fired straight up?

A 2.0 cm 2) 0 cm steel rod of length 80 cm has its left end clamped and a force of Applied perpendicular to its length at its right end.By how much does the right end shift from its original position? For steel Young's modulus is ,the bulk modulus is ,and the shear modulus is )

A uniform pressure of 21 × 105 N/m2 is applied to all six sides of a copper cube.What is the percentage change in volume of the cube? (for copper,B = 14 × 1010 N/m2)

A copper alloy wire of length 2.0 m,cross sectional area 7.1 × 10−6 m2,and Young's modulus 11 × 1010 N/m2,has a 200-kg load hung on it.What is its increase in length? (g = 9.8 m/s2)

How large a force is necessary to stretch a 2.0 mm radius steel wire (Y = 2.0 × 1011 N/m2) by 2.0%?

An ore car of mass 2000 kg rolls downhill on tracks from a mine.At the end of the tracks,10.0 m lower in elevation,is a spring with k = 400,000 N/m.How much is the spring compressed in stopping the ore car? Ignore friction.

Suppose a 0.30-kg mass on a spring that has been compressed 0.20 m has elastic potential energy of 1.0 J.How much further must the spring be compressed to triple the elastic potential energy?

To double the total energy of a simple harmonic oscillator,by what factor should the amplitude change?

I attach a 2.0-kg block to a spring that obeys Hooke's Law and supply 4.0 J of energy to stretch the spring.I release the block;it oscillates with period 0.30 s.The amplitude is:

Suppose a 0.30-kg mass on a spring that has been compressed 0.10 m has elastic potential energy of 1.5 J.What is the spring constant?

A uniform pressure of 21 × 10⁵ N/m² is applied to all six sides of a copper cube.What is the percentage change in volume of the cube? (for copper,B = 14 × 10¹⁰ N/m²)

A copper alloy wire of length 2.0 m,cross sectional area 7.1 × 10−⁶ m²,and Young's modulus 11 × 10¹⁰ N/m²,has a 200-kg load hung on it.What is its increase in length? (g = 9.8 m/s²)

How large a force is necessary to stretch a 2.0 mm radius steel wire (Y = 2.0 × 10¹¹ N/m²) by 2.0%?