YOUNG'S MODULUS

20.0 𝑘𝑔 mass is hung from a 2.0 𝑚 long vertical wire. If the wire is elongated by 3.0 𝑚𝑚, calculate the strain energy stored in the wire.

• 0.99 J

• 0.29 J

• 0.09 J

• 0.19 J

Answer explanation

The strain energy (U) in the wire can be calculated using U = 0.5 * F * x, where F is the force (weight = mg) and x is the extension. Here, F = 20.0 kg * 9.81 m/s² = 196.2 N and x = 3.0 mm = 0.003 m. Thus, U = 0.5 * 196.2 N * 0.003 m = 0.29 J.

Tags

NGSS.HS-PS3-1

NGSS.HS-PS3-2

A 10.0 𝑘𝑔 mass is attached to a 3.0 𝑚 long cable. If the cable is stretched by 2.0 𝑚𝑚, calculate the strain energy stored in the cable.

• 0.60 J

• 0.20 J

• 0.40 J

• 0.80 J

Answer explanation

The strain energy (U) in the cable can be calculated using U = 0.5 * F * x, where F is the force (weight = mg = 10 kg * 9.81 m/s²) and x is the extension (2.0 mm = 0.002 m). This gives U = 0.40 J, the correct answer.

Tags

NGSS.HS-PS3-2

A 25.0 𝑘𝑔 load is hung from a 2.5 𝑚 long elastic band. If the band elongates by 5.0 𝑚𝑚, find the elastic potential energy stored in the band.

• 1.25 J

• 1.75 J

• 0.50 J

• 1.00 J

Answer explanation

The elastic potential energy (EPE) is given by EPE = 0.5 * k * x^2. First, find the spring constant k using F = kx, where F = mg = 25.0 kg * 9.81 m/s² and x = 0.005 m. This gives k = 4905 N/m. Thus, EPE = 0.5 * 4905 * (0.005)^2 = 1.25 J.

Tags

NGSS.HS-PS3-1

A cylindrical rod with a length of 3.0 𝑚 and a cross-sectional area of 1.0 𝑐𝑚² is subjected to a tensile force of 150 𝑁. If the rod elongates by 0.3 𝑐𝑚, calculate the Young’s modulus of the material.

• 1.50 x 10⁹ Pa

• 2.50 x 10¹⁰ Pa

• 1.50 x 10¹⁰ Pa

• 4.50 x 10¹⁰ Pa

Answer explanation

Young's modulus (E) is calculated using the formula E = (F/A) / (ΔL/L). Here, F = 150 N, A = 1.0 cm² = 1.0 x 10⁻⁴ m², ΔL = 0.3 cm = 0.003 m, and L = 3.0 m. Plugging in the values gives E = 2.50 x 10¹⁰ Pa, which is choice 2.

A wire fixed to a ceiling has a length of 1.5 𝑚 and cross-sectional area 0.2 𝑐𝑚². The wire stretches by 0.15 𝑐𝑚 when a 50 𝑘𝑔 load is attached to its free end. Calculate the Young’s modulus of the wire

• 5.45 x 10¹⁰ Pa

• 3.45 x 10¹⁰ Pa

• 4.45 x 10¹⁰ Pa

• 2.45 x 10¹⁰ Pa

Answer explanation

Young's modulus (E) is calculated using the formula E = (F/A) / (ΔL/L). Here, F = 50 kg × 9.81 m/s² = 490.5 N, A = 0.2 cm² = 2 x 10⁻⁵ m², ΔL = 0.15 cm = 0.0015 m, and L = 1.5 m. Plugging in the values gives E = 2.45 x 10¹⁰ Pa.

A 50 𝑐𝑚 wire has Young’s modulus 175 𝐺𝑃𝑎 and diameter 0.25 𝑚𝑚. Calculate the force needed to elongate the wire 1.5 𝑚𝑚.

• 35.8 N

• 25.8 N

• 65.8 N

• 45.8 N

Answer explanation

To find the force, use the formula F = (Y * A * ΔL) / L, where A is the cross-sectional area. The area A = π(d/2)². Substituting values gives F = 25.8 N, confirming the correct answer is 25.8 N.

A 75 𝑐𝑚 wire has Young’s modulus 200 𝐺𝑃𝑎 and diameter 0.3 𝑚𝑚. Calculate the force required to stretch the wire 2.0 𝑚𝑚.

• 45.2 N

• 5575.2 N

• 7235.2 N

• 85.2 N

Answer explanation

To find the force, use the formula F = (Y * A * ΔL) / L, where A is the cross-sectional area. The area A = π(d/2)². Substituting values gives F = 45.2 N, confirming the correct answer is 45.2 N.

A 30.0 𝑘𝑔 mass is suspended from a 2.0 𝑚 long elastic cord. If the cord stretches by 6.0 𝑚𝑚, calculate the elastic potential energy stored in the cord.

• 0.90 J

• 0.70 J

• 0.50 J

• 0.30 J

Answer explanation

The elastic potential energy (EPE) is calculated using EPE = 0.5 * k * x^2. First, find the spring constant k using k = mg/x, where m = 30 kg, g = 9.81 m/s², and x = 0.006 m. This gives k = 4905 N/m. Then, EPE = 0.5 * 4905 * (0.006)^2 = 0.90 J.

Tags

NGSS.HS-PS3-1

NGSS.HS-PS3-2

A 12.0 𝑘𝑔 weight is attached to a 1.8 𝑚 long wire. If the wire is elongated by 2.5 𝑚𝑚, determine the strain energy stored in the wire.

• 0.35 J

• 0.25 J

• 0.15 J

• 0.45 J

Answer explanation

The strain energy (U) in the wire can be calculated using the formula U = 0.5 * F * ΔL, where F is the force (weight) and ΔL is the elongation. Here, F = 12 kg * 9.81 m/s² = 117.72 N and ΔL = 0.0025 m. Thus, U = 0.5 * 117.72 N * 0.0025 m = 0.45 J.

Tags

NGSS.HS-PS3-1

A 40 𝑐𝑚 wire has Young’s modulus 150 𝐺𝑃𝑎 and diameter 0.2 𝑚𝑚. Calculate the force needed to elongate the wire 2.0 𝑚𝑚.

• 20.5 N

• 30.5 N

• 40.5 N

• 50.5 N

Answer explanation

To find the force, use the formula F = (Y * A * ΔL) / L, where A is the cross-sectional area. The area A = π(d/2)² = π(0.0001)². Plugging in values gives F ≈ 30.5 N, confirming the correct answer.

Tags

NGSS.HS-PS2-1