05%20-%20Work,%20%20Energy%20%20and%20%20Power

05%20-%20Work,%20%20Energy%20%20and%20%20Power - 5 – WORK...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 5 – WORK, ENERGY AND POWER Page 1 5.1 Work “The product of force and displacement ( in the direction of force ), during which the force is acting, is defined as work.” When 1 N force is applied on a particle and the resulting displacement of the particle, in the direction of the force, is 1 m, the work done is defined as 1 J ( joule ). The dimensional formula of work is M 1 L 2 T - 2 . The displacement may not be in the direction of the applied force in all cases. In the figure shown, the displacement, d, makes an angle θ with the applied force, F. According to the definition of force, Work, W = force × displacement in the direction of the force = F ( d cos θ ) = ( F cos θ ) ( d ) = ( the component of force in the direction of displacement ) × ( displacement ) ( i ) For θ = π / 2, work W = 0, even if F and d are both non-zero. In uniform circular motion, the centripetal force acting on a particle is perpendicular to its displacement. Hence, the work done due to centripetal force during such a motion is zero. ( ii ) If θ < π / 2, work done is positive and is said to be done on the object by the force. ( iii ) If π / 2 < θ < π , work done is negative and is said to be done by the object against the force. 5.2 Scalar product of two vectors The scalar product of two vectors, → → B and A , also known as the dot product, is written by putting a dot ( ⋅ ) between the two vectors and is defined as: → → ⋅ B A = l → A l l → B l cos θ = A B cos θ , where θ is the angle between the two vectors. To obtain the scalar product of → → B and A , they are to be drawn from a common point, O, with the same magnitudes and directions as shown in the figure. M is the foot of perpendicular from the head of → → B to A . OM ( = A cos θ ) is the magnitude of projection of → → B on A . 5 – WORK, ENERGY AND POWER Page 2 Similarly, N is the foot of perpendicular from the head of → → A to B and ON ( = B cos θ ) is the magnitude of projection of → → A on B . ∴ → → ⋅ B A = AB cos θ = B ( A cos θ ) = ( B ) ( OM ) = ( magnitude of → B ) ( magnitude of projection of → → B on A ) or → → ⋅ B A = AB cos θ = A ( B cos θ ) = ( A ) ( ON ) = ( magnitude of → A ) ( magnitude of projection of → → A on B ) Thus, scalar product of two vectors is equal to the product of magnitude of one vector with the magnitude of projection of second vector on the direction of the first vector. The scalar product of vectors is zero if the angle between the vectors θ = π / 2 , positive if 0 ≤ θ < π / 2 and negative if π / 2 < θ ≤ π . Properties of scalar product ( 1 ) Commutative law: → → ⋅ B A = AB cos θ = B A cos θ = → → ⋅ A B Thus, scalar product of two vectors is commutative....
View Full Document

{[ snackBarMessage ]}

Page1 / 12

05%20-%20Work,%20%20Energy%20%20and%20%20Power - 5 – WORK...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online