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CHAPTER-7 Kinetic Energy and Work
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Ch 7-2,3 Kinetic Energy Energy: a scalar quantity associated with state or condition of one or more objects Kinetic Energy (K): energy associated with state of motion of an object. The faster an object moves, greater is its K K=mv 2 /2 Unit of energy (K or any type of energy) is joule (J) 1J=1 kg. m 2 /s 2
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Ch 7-4 Work Work W is energy transferred to or from an object by means of a force acting on the object. If the object is accelerated by applying a force, its kinetic energy K increases. Energy transferred to the object is positive work +W. If you decelerate the object by applying a force, you decrease its kinetic energy K. Energy transferred from the object is negative work -W.
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Ch 7-5 Work and Kinetic Energy Work done in accelerating a bead through a distance d along x-axis under a constant force F acting at an angle with respect to x-axis W= K= m(v 2 -v 0 2 )/2 but a x d=(v 2 -v 0 2 )/2 W= K=m a x d=F x d=Fcos d=F.d W= F x d=Fcos d=F.d Positive Work : Displacement along force direction Negative Work: Displacement opposite to force direction
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Ch 7-5 Work and Kinetic Energy Work-Kinetic Energy Theorem W= K = K f -K i = m(v 2 -v 0 2 )/2 K f = W+ K i v 2 =2(W+K i )/m
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Ch 5-Check-Point-1 A particle moves along x-axis. Does its kinetic energy increase or decrease, or remain the same if the particle velocity changes a) from -3 m/s to -2 m/s b) -2 m/s to + 2 m/s c) in each situation the work done on the particle is positive, negative or zero? K= m(v f 2 -v i 2 )/2 a) K= m(v f 2 -v i 2 )/2 = m/2(4-9)=- 5m/2 K is negative b) K= m*(v f 2 -v i 2 )/2 = m(4-4)/2=0 K is constant c) W is negative W is Zero
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Ch 5-Check-Point-2 The figure show four situation in which a box acts on a box while the box slides rightward a distance d across a frictionless floor. The magnitude of the forces is identical: their orientation are as shown. Rank the situation according to the work done on the box during the displacement, most positive first W = Fdcos d) W= Fd c) W=Fdcos b) W= 0 a) W = - Fdcos
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Ch 7-6 : Work Done by Gravitational Force (Constant Force) A tomato, thrown upward, is slowed down from initial velocity v 0 to v under the effect of gravitational force F g : Work W g done by the gravitational force F g in rising objects: W g =F g d cos =mgd cos = mgd (-1)= - mgd Work W g done by the gravitational force F g in falling objects : W g =F g d cos =mgd cos = mgd (+1)= + mgd
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Work done in lifting an object Work W a done by an applied force F in lifting / lowering an object through a distance d: K = K f -K i = W a + W g In lifting object is at rest in initial and final position K = K f -K i = 0 then W a = - W g = -mg d cos Lifting = 180 Lowering = 0
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Ch 7-7: Work done by a spring Force (Variable Force) Relaxed state of a spring ( Fig. a): Spring neither compressed nor extended Spring Fore F x (Restoring Forces) acts to restore the spring to its relaxed state F x =-kx (Hooke’s Law) where k is spring constant ( force constant) and x is compression or extension in the relaxed length of the spring -ve F x for +ve value of x +ve F x for -ve value of x
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Ch 7-7: Work done by a spring Force (Variable Force) Work done by a spring force W s W s = xi xf F x dx = xi xf (-kx) dx= k(x i 2 -x f 2 )/2 W s = -kx 2 /2 (if x i =0 and x f =x) Work done by an applied force W a in stretching/compressing a spring K = W a +W s If the block is initially and finally at rest then K =0 and W a = - W s
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Ch 5-Check-Point-4 Work done by spring force is W s =k(x i 2 -x f 2 )/2 a)Ws= k(9-4)/2=+5k/2 J b)Ws= k(4-9)/2=-5k/2 J c)Ws= k(4-4)/2=0 J For three situations the initial and final position respectively, along x-axis for the block is a) -3 cm, 2cm b) 2 cm, 3 cm c) -2 cm, 2 cm. In each situation the work done by the spring force on the block is positive, negative, or zero.
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Ch 7-8: Work done by a General Variable Force W= xi xf F x dx Work done by a variable force is equal to area between F(x) curve and the x-axis, between the limits x i and x f Work-Kinetic Enegy Theorem for a variable force K= K f -K i =W= xi xf F x dx
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Ch 7-9 Power Power P : Time rate of doing work of a force Average Power P avg = W/ t Instantaneous Power P= dW/dt = d/dt (Fcos dx) P = Fcos v=F.v Unit of power : 1 watt= 1 W= 1J/s
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