Chapter 9 Work Power Energy

Chapter 9 Work Power Energy Work Done 1. Work 2. Work Done by Constant Force 3. Work done by Variable Force. Work Done By Standard Forces 1. Work Done by Gravity 2. Work Done by Normal 3. Work Done by Spring 4. Work Done by Static Friction 5. Work Done by Kinetic Friction 6. Work Done by Pseudo Force Work Energy Theorem 1. Work Done by Internal Forces 2. Illustrations of Kinetic Energy and Work Energy Theorem Power 1. Illustrations of Power Potential Energy 1. Potential Energy 2. Conservative and Non-Conservative Forces 3. Central Forces 4. Gravitational Potential Energy 5. Spring Potential Energy Mechanical Energy 1. Mechanical Energy Potential Energy and Force 1. Relation between Potential Energy and Force 2. Potential Energy Curve 3. Equilibrium Vertical Circular Motion 1. Concept of Vertical Circular Motion 2. Quadrant Analysis 3. Applications of Vertical Circular Motion WORK DONE Work Work Done by a Constant Force The work performed by a constant force equals the dot product of force and displacement WF-SFScos 0 Fcos W (Fcos0)s (F₁)s Where F_{1} Component of force along displacement s * cos theta W = F(s * cos theta) = F(s_{1}) Where s_{1} = Component of displacement along force Special cases Case 1: When 0 <= theta < 90 deg cos id positive. W = 0 However, F, s, and cos & are positive. W = positive The work done by a force is positive when the applied force has a component aligned with the direction of displacement. 9(< 90 deg ) F Case II: When 90 ^ prime prime <0<=180^ prime prime . cos 0 is negative. overline F , WFscos 0 theta(> 90 deg ) W = negative However, cos e is negative. The work done by a force is negative when the applied force has a component opposite to the direction of displacement. Case III: When theta = 90 deg cos e is zero. WFscos 0=0 The work done by a force is zero when the body is displaced in a direction perpendicular to the force. overline F overline s 90 deg vec s Conclusion Angle between applied force and displacement (0) 0 <= theta < 90 deg cos e is positive. 90 deg < theta <= 180 deg, c 8 is negative. theta = 90 deg cos 8 is zero. Orientation of F and s Work done vec F vec s overline F vec s vec F Positive Negative Zero Ex. Consider two identical blocks starting from the same initial position and pushed with an equal force, but one of the blocks undergoes greater displacement. Which block experiences greater work done? Sol The final position of the blocks are shown in the figure. In this case, work done on the second block is more since its displacement is more and force and displacement are in the same direction for both the blocks. Ex. Consider two identical blocks; one being pushed and other being pulled with the same force, but acting at an angle 0 with the horizontal as shown in the figure. Displacement of both blocks is same. On which block is the work done more? Sol. The two blocks have displacement s as given in the question. For the first block, As force and displacement are along the same line, Work done, W, Fscos 0Fs For the second block, Comparing equation (1) and (2) We get W_{1} > W_{2} Work done, WFscos (Fcos0)s = FS ... (1) (2) 2 Work Done in Component Form Let a force vec F = F_{x}*i + F_{y}*j + F_{z}*k on a particle, due to which the displacement of the particle is vec s = s_{x}*i + s_{y}*j + s_{z}*k . Work done is given by the following: W = (F_{x}*i + F_{y}*j + F_{z}*k)(s_{x}*i + s_{y}*j + s_{z}*k) N = F_{x}*s_{x} + F_{y}*s_{y} + F_{x}*s_{x} W = vec F * vec S R(5, 5) Ex. A particle is displaced from point A (1, 2) to B (5, 5) by applying a constant force, vec F = (2i + 3j) * N Find the work done by the force to move the particle from point A to point B. A(1, 2) O(0, 0) [ Given, Sol Force, vec F = 2i + 3j Initial position, Final position, Displacement vector vec s = vec r 2 - vec r 1 =4 vec i +3 vec j work done by the force is given by the following: vec r_{1} = i + 2j vec r_{2} = 5i + 5j W = vec F * vec s W = (2i + 3j)(4i + 3j) W = 171 10 kg F = 10N Ex. A block of mass 10 kg is kept on a smooth horizontal surface. If a force of 10 N is applied on the block for 2 s, what is the work done by that force? Sol. Given, Mass of block (m) =10| Magnitude of force (F) = 10N Time interval (t) = 2s Let the displacement of the block in 2 seconds be s. Let a be the acceleration of the block. Applying Newton's second law of motion, we get, Using second equation of motion, we get, Work done by the force is given by the following: W = 20 J a = 10/10 = 1m * s ^ - 2 s = 0 + (1/2 * 1 * 2 ^ 2) = 2m s = ut + 1/2 * a * t ^ 2 F = ma W = Fs * cos theta = 10 * 2cos 0 deg Work Done by Multiple Forces The cumulative work performed on a particle equals the sum of the work done by all the constant forces acting upon it. Let multiple forces F_{1}, F_{2} , vec F 3 ,....... vec F n act upon a particle such that the particle undergoes a total displacement of vec s as shown in the figure. The work done is given by the following: W = vec F 1 * vec s + vec F 2 * vec s +***....+ vec F n * vec s F_{1}' vec F * P_{1} overline F_{2} 3 As the displacement of particle is constant, W =( vec F 1 + vec F 2 +***.+ vec F n )* vec s W=( vec F net )* vec s Salient features of work 1. The force's source or agent is responsible for performing the work. 2. Work done is a scalar value. 3. Displacement is measured from the point where the force is applied. 4. The joule is the Sl unit of work. 5. When multiple constant forces are at play, the work done by one force for a given displacement remains unaffected by the presence of other forces. 6 A force is independent of the reference frame, whereas its displacement varies with the reference frame, making the work done by a force frame-dependent. Ex. A block of mass M is pulled along a horizontal surface by applying a force at an angle 8 with the horizontal. The friction coefficient between the block and the surface is µ. If the block travels at a uniform velocity, find the work done by this applied force during a displacement d of the block. F M NFsin Sol mu_{2} F * cos theta Ma Given, Mass of the block = M Friction coefficient between the block and surface = mu Displacement of the block = d Let the work done by the force F be W. Since the block is moving with a uniform velocity, so the acceleration is zero. By Newton's second law of motion, vec F net =m vec a =0 Balancing force in the horizontal direction Fcos0 theta = mu*N vec Sigma*F x =0 Balancing force in the vertical direction, Substituting the value of N from equation (2) in equation (1), N + F * sin theta = Mg F = (mu*Mg)/(cos theta * mu * sin theta) F * cos theta = mu*Mg - mu*F * sin theta F\{cos theta + mu * sin theta\} = mu*Mg F * cos theta = mu\{Mg - F * sin theta\} Work done by the force is given by the following: vec Sigma*F y =0 W = (mu*Mgd * cos theta)/(cos theta + mu * sin theta) W = F * d * cos theta = (mu*M_{R})/(cos theta + mu * sin theta) * d * cos theta (1) ... (2) Work Done by a Variable Force The cumulative work done equals the summation of infinitesimal work done over infinitesimal displacements. W = integrate 1 dW The work done by a force in transporting a particle from point A to point B is expressed as follows: W= int A ^ B vec F * d vec r = integrate F * cos theta dr from A to B Note: When a force's magnitude or direction varies with time or position, it is termed as a variable force. 4 Work Done by a Variable Force In Rectangular Form Let a force, F = F_{x}*i + F_{y}*j + F_{x}*k result in the displacement of the particle, d vec r = dxi + dyj + dzk Consider A and B as the starting and ending positions, respectively, of the particle. Work done by the force is given by, dW= vec F * d vec r = F_{x} * dx + F_{y} * dy + F_{x}*dz W = integrate 1 dW W= int A ^ B (F x dx+ F_{y} * dy + F_{z}*dz ) W = integrate F_{x} dx from x_{A} to x_{B} + integrate F_{y} dy from y_{A} to y_{B} + integrate F_{x} dz from x_{A} to z_{B} Area Under the Force vs Displacement Graph The area under the graph of force versus displacement provides the measure of work done by that force over the given displacement. The total work done is the accumulation infinitesimal work over infinitesimal displacements along the track. W= int r 1 ^ r t vec F * d vec r df F * (r) F(t) Ex. A force vec F = (0.5x + 10) iacts on a particle. Here, vec F is in newton and x is in metre. Calculate the work done by the force during the displacement of the particle from x = 0 m x = 2n Sol Given, Force, vec F = (0.5x + 10) * i Infinitesimal work done by the force is given by the following dW= vec F * d vec x = (0.5x + 10) * i . (dx)i = 0.5xdx + 10dx Total work done by the force is given by the following: W = ∫ dw = (0.5xdx + 10dx) Alternate way F(0) = 10N The force vs displacement graph is as shown in the figure. Area under the force vs displacement curve gives the work done. 10 F(2) = 11N W = 1/2 * \{(10 + 0) + (11 + 0)\}(2 + 0) F = 0.5x + 10N W= 0.5 2 [x ^ 2] 0 ^ 2 +10[x] 0 ^ 2 W = 21I w = 21l F \{N\} 11 x(m) Ex. A force, vec F = - k * (yi + xj) acts on a particle moving in the x-y plane, where, k is a positive constant. The particle starts from the origin and is taken along the positive x axis to the point (a, 0), and then perpendicularly to the point (a, a), Calculate the total work done on the particle by the force F. C(0,a) B(a, a) overline overline F =-k(y overline i +a overline j ) vec F =-k(x vec j ) A(a, 0) Sol Given, Force, vec F =-k( yi +x hat j ) Total work done Work done in taking the particle from O to A+ Work done in taking the particle from A to B. W=W OA +W AB 5