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Unit 05a

Unit 05a. “Work, Power and Energy” Work - Kinetic Energy Theorem. The “Work-Kinetic Energy Theorem” states …. The work done on an object is equal to the change in the object’s kinetic energy. The work done is equal to the energy used! More work done, more change in energy!.

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Unit 05a

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  1. Unit 05a “Work, Power and Energy” Work - Kinetic Energy Theorem

  2. The “Work-Kinetic Energy Theorem” states … • The work done on an object is equal to the change in the object’s kinetic energy. • The work done is equal to the energy used! • More work done, more change in energy!

  3. This makes sense because … • A force can • Speed up • Slow down • Change direction • The more work you do, the more the object will speed up or slow down! • A change in velocity is a change in kinetic energy! The force of the parachute slows the man down over a certain distance. Work changes kinetic energy!

  4. Work- Kinetic Energy Equation W = ΔKE W = KEf – KEi Fd = ½mVf2 – ½mVi2

  5. Work- Kinetic Energy Equation W = ΔKE Fd ½mVf2 ½mVi2 KEf – KEi

  6. Problem Solving1. Medium (-) “Step by Step A 500kg car is pushed from rest with a force of 400N, if the car reaches a final velocity of 10m/s, how far is it pushed? m = 500kg Vi =0m/s F = 400N Vf = 10m/s d = ? c. Find the work done on the car. d. Find the distance the car moves. W = KEf - KEi • a. Find the initial kinetic energy. • Find the final kinetic energy. W = 25000J – 0J KEi = ½ mVi2 W = 25000J KEi = 0 J KEf = ½ mVf2 W = Fd KEf = ½ (500kg)(10m/s)2 25000J = (400N)d KEf = ½ (500kg)(100m2/s2) 62.5m = d KEf = 25000 J

  7. 2. Medium A 30kg baby carriage is pushed from rest for 2.5m with a force of 2900N, how far was it pushed if it reached a final velocity of 4m/s? Fd = ½mVf2 – ½mVi2 m = 30kg 2900N(d) = ½(30kg)(4m/s)2 – ½(30kg)(0m/s)2 Vi = 0m/s 2900N(d) = ½(30kg)(16m2/s2) – 0 J F = 2900N d = ? 2900N(d) = 240kgm2/s2 – 0 J Vf = 4m/s 2900N(d) = 240J d = 0.0828m

  8. 3. Medium A 65kg boy is skating with a velocity of 5m/s, if he pushes himself to slow down for a distance of 4m and reaches a final velocity of 2m/s, what force did he apply? Fd = ½mVf2 – ½mVi2 m = 65kg F(4m) = ½(65kg)(2m/s)2 – ½(65kg)(5m/s)2 Vi = 5m/s F(4m) = ½(65kg)(4m2/s2)– ½(65kg)(25m2/s2) d = 4m F(4m) = 130kgm2/s2 – 813kgm2/s2 Vf = 2m/s F(4m)= -683J F= ? F = 171 N

  9. 4. Medium + A 45kg shopping cart is pushed from rest for 5m with a force of 300N, how fast is it going after? Fd = ½mVf2 – ½mVi2 m = 45kg (300N)(5m)=½(45kg)Vf2 -(45kg)(0m/s)2 Vi = 0m/s 1500J = 22.5kgVf2 – 0 J d = 5m 1500J = 22.5kgVf2 F = 300N 66.6m2/s2 = Vf2 Vf = ? 8.16m/s = Vf

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