PH20Unit3Assignment3

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Jan 9, 2024

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Energy and Work: Unit 3 Lesson 3 Assignment KEY Total /47 1. When a force of 18.0 N is applied to an object, the work done is 454 J. What distance does the object move? W=Fd F=18 N W=454J d=? 454 =18d d=25.2 m /2 2. A force applied to a 25 kg mass causes an acceleration of 3.5 m/s 2 over 25 m. The work done on the object is m=25 kg a=3.5 m/s2 d=25 m W=? F=ma = 25*3.5 = 87.5 N W=fm=87.5 *25 W=2.2*103 /3 3. A force of 52 N is applied to a 12 kg mass resting on a bench where the force of sliding friction is 15 N. The work done in overcoming friction while moving the object 24 m is F=52 N M=12 kg Ff=15 N d=24m W=?

Fr=52 - 15 =37N W=37N*24m=8.9*102 J W=907.8J /3

Use the information to answer the question. A force of 424 N is applied to an 80.0 kg box by pulling on a rope that makes an angle θ with a highly polished floor as shown. 4. The box is pulled 12.0 m across the floor at constant speed. If the work done is 4.61 × 10 3 J, what angle does the rope makes with the floor? W=4.61×10^3J Angle=? W=Fdcosx 4.61×10^3 = 424*12*cosx x= 25.0 ͦ /2 Use the information to answer the question A 160 kg carton is pulled up a 3.00 m and low-friction plank onto the back of a truck as shown. 5. If the back of the truck is 1.25 m above the ground, what force is required to move the carton at a constant speed up the plank and what is the work done? H=1.25m m=160kg d=3m F=? F=ma

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F =160*9.81 =1569.6 N W=? W=fdcosx x=? x=sin-1(1.25/3) =24.62431835° W=1569.6*3*cos24.62431835° W= 4.28*10^3J /4 6.The change in the kinetic energy of an object is equal to … The net work done on it /2 7. If an object weighing 196 N has a kinetic energy of 950 J, what is its velocity? velocity? Ek=mv^2/2 m=196/9.81 =19.97961264 kg 950=19.97961264 *v2 /2 v=9.75 m/s /3 8. The gravitational potential energy of an object is dependent on … Mass of object and how high it is /2 Use the diagram and information to answer the question. A 50.0 kg carton is dragged up a low-friction plank onto the back of the truck. The length

of the plank is 3.00 m and it makes an angle of 15.0º with the ground. 9. With respect to the bottom of the plank, the gravitational potential energy of the carton on the truck is m=50kg. h=? sin15= opp/hyp =h/3 h=0.7764571353m GPE=? mgh =50*9.81*0.77645713 = 381 J /3 10. Calculate the gravitational potential energy of an 15.8 kg object sitting on a table 2.10 m above the floor. m=15.8 kg h=2.10 m GPE=mgh GPE= 15.8*9.81*2.10 = 325 J /3 12. If a force of 256 N is applied to a spring, the spring is stretched a distance of 9.50 cm from its equilibrium position. What is the elastic constant of this spring? /3 F=256 N x=9.5 cm Fspring= -kx 256= k*9.5

k=26.9 N/cm 13. A spring with an elastic constant of is stretched to a position 7.50 cm from its equilibrium position. What is the increase in the energy stored in this spring if it is stretched another 5.00 cm? /3 k=4.50*10^3N/m x=7.50cm = 0.075m Ep=kx^2/2 Ep=4.50*10^3*0.075^2/2 =12.65625 J Increase in energy: Ep=4.50*10^30.075+0.05^2/2 =35.15625 J =35.1565 - 12.656 = 22.5 J 14. A child, initially at rest at the top of 3.55 m high slide, slides down to a height of 0.480 m. At the bottom of the slide, the child has a speed of 2.45 m/s. If the mass of the child is 34.5 kg, what is the change in mechanical energy of the child? /3 Em=Ep+Ek Ep=mgh =34.5*9.81*3.55 =1201.47975 J Initial: Em=1201.47975 J Change ME: Ek=mv^2/2=34.5*2.45^2/2 =103.543125 J Ep=mgh =34.5*9.81*0.480=162.45361 Em=103.543125+162.4536=265.996725 J

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1201.47975 J-265.996725J = 935 J The diagram shows a skateboarder starting to roll down a hill. Assume that the skateboard rolls with negligible friction. Diagram is not to scale. 15. Refer to the above diagram to answer this question. A skateboarder and his skateboard have a combined mass of 76.4 kg. The skateboarder passes a point P that is 3.90 m above bottom of the hill with a speed of 1.25 m/s. What is the skateboarder's kinetic energy as he passes another point Q that is 1.15 m above the bottom of the hill? /3 m=76.4 kg Point P: Ep=76.4 x 9.81 x 3.9 =2922.9876J Ek=76.4 x 1.252 /2 =59.6875 J Em=2922.9876 + 59.6875 = 2982.6751 J point Q: Ep=mgh =76.4 x 9.81 x 1.15

=861.9066 J Ek=? =2982.6751 J - 861.9066 J = 2120.7685J =2.12 x 10^J 16. Define power. Power is the rate at which energy is transferred and rate at which work is done. Can be written as p=e/t where p= power(watt) e=energy transferred (J) done t=time(S) /2 17. If a 70.0 kg student runs up a flight of stairs 15.0 m high in 25.0 s, what is the average power? P=E/t E=mgh =70 x 9.81 x 15 =10300.5 J P=10300.5/25 P=412 W /3 18. A 65.0 kg student climbs a set of stairs in 25.0 s with a power output of 95 W. What is the vertical height of the stairs? M=65kg T=25s P=95w H=? P=mgh/t 95=65kg x 9.81 x h/25 H=3.7m /3

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