#### Dr- Sonia Reda Dr- Sonia Reda chapter 7 Kinetic energy and Work 7.2 What is energy 7.3 Kinetic energy 7.4 Work 7.5 Work and kinetic Energy 7.6 Work done by the.

Download Report#### Transcript Dr- Sonia Reda Dr- Sonia Reda chapter 7 Kinetic energy and Work 7.2 What is energy 7.3 Kinetic energy 7.4 Work 7.5 Work and kinetic Energy 7.6 Work done by the.

Dr- Sonia Reda Dr- Sonia Reda chapter 7 Kinetic energy and Work 7.2 What is energy 7.3 Kinetic energy 7.4 Work 7.5 Work and kinetic Energy 7.6 Work done by the gravitational force 7.7 Work done by a Spring force 7.8 Power Outline Chapter 7 Work and Kinetic energy Work done by a net force results in kinetic energy Some examples: gravity, spring, friction What is Energy? The term energy is so broad that a clear definition is difficult to write. Technically, Energy is a scalar quantity associated with the state (or condition) of one or more objects. However, this definition is too vague to be of help to us now. Dr- Sonia Reda Kinetic Energy Kinetic energy K is energy associated with the state of motion of an object. For an object of mass m whose speed v is well below the speed of light, Kinetic energy K is: Unit for Kinetic energy is: Kinetic energy is a scalar quantity. Dr- Sonia Reda Work Work W is energy transferred to or from an object by means of a force acting on the object. Energy transferred to the object is positive work, Energy transferred from the object is negative work. Dr- Sonia Reda Properties of Work Only the force component along the object’s displacement will contribute to work. The force component perpendicular to the displacement does zero work. A force does positive work when it has a vector component in the same direction displacement, A force does negative work when it has a vector component in the opposite direction. Work is a scalar quantity. Dr- Sonia Reda Finding an Expression for Work we can use Eq. 2-16 to write, for components along the x axis, v2 =vo2 + 2axd By multiplying the above Eq with ½ m Dr- Sonia Reda Finding an Expression for Work Dr- Sonia Reda Kinetic Energy Work-Kinetic Energy Theorem Change in KE work done by all forces DK Dw Dr- Sonia Reda Work-Kinetic Energy Theorem Vector sum of all forces acting on the body SF w x F .dx xf i xx ma.dx f i x x i f dv m .dx dt x x i f dx m .dv dt xi xf x m v .dv m[1/ 2v ] vf vi 2 vf vi = 1/2mvf2 – 1/2mvi2 = Kf - Ki = DK Work done by net force = change in KE Dr- Sonia Reda Checkpoint 1 A particle moves along an x axis. Does the kinetic energy of the particle increase, decrease, or remain the same if the particle’s velocity changes (a) from −3 m/s to −2 m/s and (b) from −2 m/s to 2 m/s? (c) In each situation, is the work done on the particle positive, negative, or zero? Dr- Sonia Reda Example 7-3 During a storm, a crate of crepe is sliding across a slick, oily parking lot through a displacement while a steady wind pushes against the crate with a force . The situation and coordinate axes are shown in Fig. 7-5. . (a) How much work does this force do on the crate during the displacement? Dr- Sonia Reda (a) How much work does this force from the wind do on the crate during the displacement? SOLUTION: Work done by the wind force on crate : W F d (2.0 N) ˆi ( 6.0 N) ˆj ( 3.0 m) i (2.0 N) ( 3.0 m) ˆi ˆi ( 6.0 N) ( 3.0 m) ˆj ˆi ( 6.0 J) (1) 0 6.0 J The wind force does negative work, i.e. kinetic energy is taken out of the crate. Dr- Sonia Reda (b) If the crate has a kinetic energy of 10 J at the beginning of displacement d , what is its kinetic energy at the end of d ? SOLUTION: K f K i W 10 J ( 6.0 J) 4.0 J Dr- Sonia Reda Gravitation and work Work done by me (take down as +ve) h F mg Lift mass m with constant velocity = F.(-h) = -mg(-h) = mgh Work done by gravity = mg.(-h) = -mgh ________ Total work by ALL forces (DW) = 0 =DK Work done by ALL forces = change in KE DW = DK What happens if I let go? Dr- Sonia Reda Work Done by a Spring Force The spring force given by Hooke’s Law: spring x F k Dx The work done by spring force: 1 1 W spring ( kx22 kx12 ) 2 2 Dr- Sonia Reda Compressing a spring Compress a spring by an amount x F -kx x Work done by me Fdx = kxdx = 1/2kx2 Work done by spring -kxdx =-1/2kx2 Total work done (DW) = 0 =DK What happens if I let go? Dr- Sonia Reda Moving a block against friction at constant velocity f F d Work done by me = F.d Work done by friction = -f.d = -F.d Total work done What happens if I let go? = 0 NOTHING!! Gravity and spring forces are Conservative Friction is NOT!! Dr- Sonia Reda Sample Problem 7-8 In Fig. 7-11, a cumin canister of mass m = 0.40 kg slides across a horizontal frictionless counter with speed v = 0.50 m/s. It then runs into and compresses a spring of spring constant k = 750 N/m. When the canister is momentarily stopped by the spring, by what distance d is the spring compressed? Dr- Sonia Reda SOLUTION: We assume the spring is massless. Work done by the spring on the canister is negative. This work is : WS 12 kd2 Kinetic energy change of the canister is : k f ki mv 1 2 Therefore, 2 12 kd2 12 mv2 dv m 0.40 kg (0.50 m / s) k 750 N / m 1.2 x 10 2 m 1.2 cm Dr- Sonia Reda Power The rate at which work is done by a force is called the power. The average power due to the work done by a force during that time interval as We define the instantaneous power P as the instantaneous rate of doing work, so that W = F . Δx Dr- Sonia Reda The units of power Dr- Sonia Reda Sample Problem 7-10 Figure 7-14 shows constant forces F1 and F2 acting on a box as the box slides rightward F across a frictionless floor. Force 1 is horizontal, with magnitude 2.0 N; force F2 is angled upward by 60° to the floor and has magnitude 4.0 N. The speed v of the box at a certain instant is 3.0 m/s. Dr- Sonia Reda (a) What is the power due to each force acting on the box at that instant, and what is the net power? Is the net power changing at that instant? SOLUTION: P1 F1v cos 180 (2.0 N ) (3.0 m / s ) cos 180 6.0 W P2 F2 v cos 60 (4.0 N ) (3.0 m / s ) cos 60 6.0 W Pnet P1 P2 0 The kinetic energy of the box is not changing. The speed of the box remains at 3 m/s. The net power does not change. Dr- Sonia Reda (b) If the magnitude of F2 is, instead, 6.0 N, what now is the net power, and is it changing? SOLUTION: P2 F2 v cos 60 (6.0 N) (3.0 m / s) cos 60 9.0 W Pnet P1 P2 6.0 W 9.0 W 3.0 W There is a net rate of transfer of energy to the box. The kinetic energy of the box increases. The net power also increases. Dr- Sonia Reda Dr- Sonia Reda