Ch 4 and 5 HW key - Chapter 4 1. KE= mv= * 2 * 20= 400 J 2....

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Chapter 4 1. KE= ½ mv²= ½ * 2 * 20²= 400 J 2. Workdone= KE+ PE= 1/2mv² + mgh= ½* 50* 1²+ 50*9.81*3.5= 1741.75 J 3. a.yes, positive(moving the same direction as the displacement) b .concentric contraction. c. yes, negative (moving the different direction as the displacement) d. eccentric contraction. 4. F= 60N, m= 0.15kg, d=2m a. U=Fd= 60*2= 120 J b . positive , force act the same direction as displacement c.Vi=0, Vf=? U=ΔE= ΔKE=1/2m(vf²-vi²)=1/2mVf²=1/2* 0.15* V f²=120J-> Vf= 40 m/s 5. V=40m/s, m=0.15kg, d=8cm=0.08m a. KE= ½ mv²= ½*0.15*40² = 120J b.Ef=Ei- > 120J c. negative , direction of force and the direction of the ball is opposite d.U=Fd-> F= U/d= 120/ 0.08= 1500N 6. KE= ½ mv² 5kg roll 4m/s-> KE=1/2 *5* 4²= 40J <- more energy 6kg roll 3m/s-> KE=1/2 *6*3²=27J
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7. t= 0.5, d=2m. m=100kg P= u/t= PE/t=mgh/t=( 100*10*2)/ 0.5= 4000W 8. P= u/t= PE/t=mgh/t, g and t is the same, so if the product of m*h is greater, then the power’s greater. m1h1= 60*60=1200
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Ch 4 and 5 HW key - Chapter 4 1. KE= mv= * 2 * 20= 400 J 2....

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