(a) Define Young's modulus (b) When a force of 50... | WAEC SSCE - Physics - 2004

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(a) Define Young's modulus

(b) When a force of 50 N is applied to the free end of an elastic cord, an extension of 4 cm is produced in the cord. Calculate the work done on the cord.

Answer

(a) Young's modulus: the ratio of the tensile stress to the tensile strain of a material, within the limit of proportionality (elastic limit). That is:

\[ E = \frac{\text{tensile stress}}{\text{tensile strain}} = \frac{F/A}{e/l} = \frac{Fl}{Ae} \]

where \(F\) is the applied force, \(A\) the cross-sectional area, \(e\) the extension and \(l\) the original length.

(b) Force \(F = 50\ \text{N}\), extension \(e = 4\ \text{cm} = 0.04\ \text{m}\). For an elastic (Hooke's-law) cord the force builds up uniformly from 0 to \(F\), so the work done (energy stored) is:

\[ W = \frac{1}{2} F e = \frac{1}{2} \times 50 \times 0.04 \] \[ W = 1.0\ \text{J} \]

The work done on the cord is \(1.0\ \text{J}\).

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Answer Details

(a) Young's modulus: the ratio of the tensile stress to the tensile strain of a material, within the limit of proportionality (elastic limit). That is:

\[ E = \frac{\text{tensile stress}}{\text{tensile strain}} = \frac{F/A}{e/l} = \frac{Fl}{Ae} \]

where \(F\) is the applied force, \(A\) the cross-sectional area, \(e\) the extension and \(l\) the original length.

\[ W = \frac{1}{2} F e = \frac{1}{2} \times 50 \times 0.04 \] \[ W = 1.0\ \text{J} \]

The work done on the cord is \(1.0\ \text{J}\).

Read lesson note on Work, Energy & Power (JAMB)

Read lesson note on Elasticity (JAMB)

Work, Energy & Power

Elasticity

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(a) Define Young's modulus (b) When a force of 50 N is applied to the free end of an elastic cord, an extension of 4 cm is produced in the cord. Calculate t...

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(a) Define Young's modulus (b) When a force of 50 N is applied to the free end of an elastic cord, an extension of 4 cm is produced in the cord. Calculate t...

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