Question

Problem 1. When a mass of 1458g is hung from a spring the spring stretches 6cm. What is the
spring constant? Express your answer in N/m.
Problem 2. The spring constant for a certain spring is found to be 6000N/m. What force is required
to compress the spring for a distance of 10 cm?
Problem 3. A 75-kg mass is suspended by means of a cable that has a diameter of 8mm. What is the
stress?
Problem 4. A 50-cm length of wire is stretched to a new length of 55.1 cm what is the strain?
Explain the answer in percent form.
2% each problem
Problem 5. Young’s Modulus for a certain rod is 9x1011 Pa. What strain will be produced by a
tensile stress of 750MPa? Explain your answer.
Problem 6. A patient’s leg was put into traction, stretching the
femur from a length of 0.46 m to 0.461 m. The femur has a diameter
of 3.15 cm. With the knowledge that bone has a Young’s modulus
of ~ 1.6 x 1010Pa in tension, what force was used to stretch the
femur?

Problem 7. A mass of 2.00 kg is supported by a copper wire of length 4.00 m and diameter 4.00
mm. Determine (a) the stress in the wire and (b) the elongation of the wire.

Answer 1

such a lameeeeeeeee questionn

Answer 2

Answer:

Multiple answers are there, please refer to the explanation

Explanation:

• 1. Mass = 1458g= 1.458 kg, $g=10 m/s^2$

force due to gravity = mg = 1.458*10= 14.58N

Force due to spring = kx, where k is spring constant and x is the stretching, x= 6 cm=0.06 m

in equilibrium  mg = kx

⇒ k = $\frac{mg}{x}=\frac{14.58}{0.06}=$

• 2. k is given , k= 6000N/m

x= 10cm = 0.1 m

F= kx = 6000*0.1 =600N

• 3.Mass = 75 kg, diameter of the cable = 8mm, ⇒ radius = 4mm

area = $\pi r^2= 3.24 * (4*10^{-3})^2= 51.84*10^{-6}$

Stress = Force/Area

$\implies \frac{mg}{A}=\frac{750}{51.84 \times 10^{-6}} =1.51 \times 10^6$$N/m^2$

• 4. Strain = $\frac{\Delta L}{L}$

$\Delta L= 55.1-50 =5.1 cm$

⇒Strain=$\frac{5.1}{50}=0.102$

• 5.Young's modulus of rod = $9\times 10^{11} Pa$

tensile stress =$750Mpa= 750\times 10^6 Pa$

we know that young's modulus is given  as the ration of stress and strain

$Y= \frac{stress}{strain}\\\implies strain = \frac{stress}{Y}=\frac{750*10^6}{9*10^{11}} =83.33\times 10^{-3}$

• 6. Stretching the femur bone from 0.46 to 0.461$\implies \Delta L= 0.001$

diameter of femur bone = 3.15 cm $area = 3.14\times 1.575^2=7.78cm^2=7.78\times 10^{-4}m^2$

$Y= 1.6\times 10^{10} Pa$

We know that$Y=\frac{F/A}{\Delta L/L} \\\implies F=Y*(\frac{\Delta L}{L} )A= 1.6*10^{10}*\frac{0.001}{0.46}*7.78\times 10^{-4}=27060. 8N$

• 7. mass= 2 kg

copper wire length = 4m

diameter = 4 mm, radius = 2mm, area= $\pi \times 4\times 10\times 10^{-6}m^2=12.56* 10^{-6}m^2$

Stress = F/A= $\frac{mg}{A}=\frac{200}{12.56*10^{-6}}=15.92\times 10^{6}Pa$

Youngs modulus of copper = $121*10^9 Pa$

$Y=\frac{F/A}{\Delta L/L} \\\implies \Delta L=(\frac{FL}{AY} )= 0.00013 cm$, which is the elongation of wire.