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Physics 111: Mechanics Lecture 6 Dale Gary NJIT Physics Department Energy Energy and Mechanical Energy Work Kinetic Energy Work and Kinetic Energy The Scalar Product of Two Vectors May 22, 2017 Why Energy? Why do we need a concept of energy? The energy approach to describing motion is particularly useful when Newton’s Laws are difficult or impossible to use Energy is a scalar quantity. It does not have a direction associated with it May 22, 2017 What is Energy? Energy is a property of the state of a system, not a property of individual objects: we have to broaden our view. Some forms of energy: Mechanical: Kinetic energy (associated with motion, within system) Potential energy (associated with position, within system) Chemical Electromagnetic Nuclear Energy is conserved. It can be transferred from one object to another or change in form, but cannot be created or destroyed May 22, 2017 Kinetic Energy Kinetic Energy is energy associated with the state of motion of an object For an object moving with a speed of v 1 2 KE mv 2 SI unit: joule (J) 1 joule = 1 J = 1 kg m2/s2 May 22, 2017 Why 1 2 KE mv 2 ? May 22, 2017 Work W 1 2 1 2 mv mv0 Fx x Start with 2 2 Work “W” Work provides a link between force and energy Work done on an object is transferred to/from it If W > 0, energy added: “transferred to the object” If W < 0, energy taken away: “transferred from the object” May 22, 2017 Definition of Work W The work, W, done by a constant force on an object is defined as the product of the component of the force along the direction of displacement and the magnitude of the displacement W ( F cos q ) x F is the magnitude of the force Δ x is the magnitude of the object’s displacement q is the angle between F and x May 22, 2017 Work Unit This gives no information about the time it took for the displacement to occur the velocity or acceleration of the object Work is a scalar quantity SI Unit Newton • meter = Joule 1 2 1 2 mv mv0 ( F cos q )x 2 2 N•m=J J = kg • m2 / s2 = ( kg • m / s2 ) • m W ( F cos q ) x May 22, 2017 Work: + or -? Work can be positive, negative, or zero. The sign of the work depends on the direction of the force relative to the displacement W ( F cos q ) x Work Work Work Work Work positive: W > 0 if 90°> q > 0° negative: W < 0 if 180°> q > 90° zero: W = 0 if q = 90° maximum if q = 0° minimum if q = 180° May 22, 2017 Example: When Work is Zero A man carries a bucket of water horizontally at constant velocity. The force does no work on the bucket Displacement is horizontal Force is vertical cos 90° = 0 W ( F cos q ) x May 22, 2017 Example: Work Can Be Positive or Negative Work is positive when lifting the box Work would be negative if lowering the box The force would still be upward, but the displacement would be downward May 22, 2017 Work Done by a Constant Force The work W done on a system by an agent exerting a constant force on the system is the product of the magnitude F of the force, the magnitude Δr of the displacement of the point of application of the force, and cosθ, where θ is the angle between the force and displacement vectors: F F r r II I WII Fr WI 0 F F W F r Fr cosq r r III WIII Fr IV WIV Fr cos q May 22, 2017 Work and Force An Eskimo pulls a sled as shown. The total mass of the sled is 50.0 kg, and he exerts a force of 1.20 × 102 N on the sled by pulling on the rope. How much work does he do on the sled if θ = 30°and he pulls the sled 5.0 m ? W ( F cos q )x (1.20 10 2 N )(cos 30 )(5.0m) 5.2 10 2 J May 22, 2017 Work Done by Multiple Forces If more than one force acts on an object, then the total work is equal to the algebraic sum of the work done by the individual forces Wnet Wby individual forces Remember work is a scalar, so this is the algebraic sum Wnet Wg WN WF ( F cos q )r May 22, 2017 Work and Multiple Forces Suppose µk = 0.200, How much work done on the sled by friction, and the net work if θ = 30° and he pulls the sled 5.0 m ? Fnet , y N mg F sin q 0 N mg F sin q W fric ( f k cos180 )x f k x k Nx k (mg F sin q )x (0.200)(50.0kg 9.8m / s 2 Wnet WF W fric WN Wg 1.2 10 2 N sin 30 )(5.0m) 5.2 10 2 J 4.3 10 2 J 0 0 4.3 10 2 J 90.0 J May 22, 2017 Kinetic Energy Kinetic energy associated with the motion of an object 1 2 KE 2 mv Scalar quantity with the same unit as work Work is related to kinetic energy 1 2 1 2 mv mv0 Fnet x 2 2 Wnet KEf KEi KE May 22, 2017 May 22, 2017 Work-Kinetic Energy Theorem When work is done by a net force on an object and the only change in the object is its speed, the work done is equal to the change in the object’s kinetic energy Speed will increase if work is positive Speed will decrease if work is negative Wnet KEf KEi KE Wnet 1 2 1 2 mv mv0 2 2 May 22, 2017 Work and Kinetic Energy The driver of a 1.00103 kg car traveling on the interstate at 35.0 m/s slam on his brakes to avoid hitting a second vehicle in front of him, which had come to rest because of congestion ahead. After the breaks are applied, a constant friction force of 8.00103 N acts on the car. Ignore air resistance. (a) At what minimum distance should the brakes be applied to avoid a collision with the other vehicle? (b) If the distance between the vehicles is initially only 30.0 m, at what speed would the collisions occur? May 22, 2017 Work and Kinetic Energy 3 3 (a) We know v0 35.0m / s, v 0, m 1.00 10 kg, f k 8.00 10 N Find the minimum necessary stopping distance Wnet 1 2 1 2 W fric Wg WN W fric mv f mvi 2 2 1 2 f k x 0 mv0 2 1 (8.00 10 N )x (1.00 103 kg)(35.0m / s) 2 2 3 x 76.6m May 22, 2017 Work and Kinetic Energy (b) We know x 30.0m, v0 35.0m / s, m 1.00 103 kg, f k 8.00 103 N Find the speed at impact. Write down the work-energy theorem: 1 1 Wnet W fric f k x mv2f mvi2 2 2 2 v 2f v02 f k x m 2 3 2 2 v 2f (35m / s) 2 ( )( 8 . 00 10 N )( 30 m ) 745 m / s 1.00 103 kg v f 27.3m / s v0 35.0m / s, v 0, m 1.00 103 kg, f k 8.00 103 N May 22, 2017 Work Done By a Spring Spring force Fx kx May 22, 2017 Spring at Equilibrium F =0 May 22, 2017 Spring Compressed May 22, 2017 xf lim Fx x Fx dx x 0 xi xf xi xf xf xi xi 0 kx dx 12 kx 2 W Fx dx kx dx xmax xf W kx dx 12 kxi2 12 kx 2f xi Work done by spring on block Fig. 7.9, p. 173 Measuring Spring Constant Start with spring at its natural equilibrium length. Hang a mass on spring and let it hang to distance d (stationary) From Fx kx mg 0 mg k d so can get spring constant. May 22, 2017 Scalar (Dot) Product of 2 Vectors The scalar product of two vectors is written as A B A B A B cos q It is also called the dot product q is the angle between A and B Applied to work, this means W F r cosq F r May 22, 2017 Dot Product The dot product says something about how parallel two vectors are. The dot product (scalar product) of two vectors can be thought of as the projection of one onto the direction of the other. A B AB cos q A iˆ A cos q Ax Components A B Ax Bx Ay By Az Bz B ( A cos q ) B A A( B cos q ) q May 22, 2017 Projection of a Vector: Dot Product The dot product says something about how parallel two vectors are. The dot product (scalar product) of two vectors can be thought of as the projection of one onto the direction of the other. A B AB cos q A iˆ A cos q Ax Components A B Ax Bx Ay By Az Bz iˆ ˆj 0; iˆ kˆ 0; ˆj kˆ 0 iˆ iˆ 1; ˆj ˆj 1; kˆ kˆ 1 B Projection is zero p/2 A May 22, 2017 Derivation A How do we show that B Ax Bx Ay By Az Bz ? Start with A A iˆ A ˆj A kˆ x y z B Bxiˆ By ˆj Bz kˆ Then A B ( A iˆ A ˆj A kˆ) ( B iˆ B ˆj B kˆ) x y z x y z Axiˆ ( Bxiˆ By ˆj Bz kˆ) Ay ˆj ( Bxiˆ By ˆj Bz kˆ) Az kˆ ( Bxiˆ By ˆj Bz kˆ) But So iˆ ˆj 0; iˆ kˆ 0; ˆj kˆ 0 iˆ iˆ 1; ˆj ˆj 1; kˆ kˆ 1 A B Axiˆ Bxiˆ Ay ˆj By ˆj Az kˆ Bz kˆ Ax Bx Ay By Az Bz May 22, 2017 Scalar Product A 2iˆ 3 ˆj and B iˆ 2 ˆj The vectors Determine the scalar product A B ? A B Ax Bx Ay By 2 (-1) 3 2 -2 6 4 Find the angle θ between these two vectors A Ax2 Ay2 2 2 32 13 A B 4 4 cos q AB 13 5 65 4 q cos 1 60.3 65 B Bx2 B y2 (1) 2 2 2 5 May 22, 2017