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Chapter 3 HW key copy

# Chapter 3 HW key copy - Chapter 3 1.m=10kg->...

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Chapter 3 1.m=10kg-> W=mg=9.81*10=98.1N (gravity is pulling the weight down with a force of 98.1N) You are pulling up with a force of 108.1 N. Combine the upward and downward forces to get a resultant force of 10 and put that into F=ma ΣF=ma= -> 10 = 10a= 1m/s/s 2.Σf=ma=1.0* 5.0= 5N in the direction of the acceleration (because the acceleration was horizontal we did not need to consider gravity. 3.a. according to Netwon’s 3 rd Law -> F Tonya = F Nancy = 1000N (when 2 objects collide the force acting on each is the same, this goes for everything from hockey sticks and pucks, a boxing glove and a chin or an athlete tackling another athlete) b. ΣF= ma=1000 -> a= 1000/m mt= 60kg, mN=50kg Tonya: 1000/60= 16.67/s/s, Nancy: 1000/50= -20m/s/s (opposite direction) (although the forces of two objects that collide are the same, their effects are different depending in part on the objects mass. The object with the least mass is affected more than the object with greater mass. This is why Nancy is accelerated more than Tonya by the same force) c. mt= 60kg, mN=50kg, uT=5m/s, uN= -6m/s mTuT+ mNuN= (mT+mN)v 60*5+ 50*(-6)= (60+50) v -> v= 0 m/s 4.R=W=mg= 15*9.81= 147.15N (since it weighs 147.15 N applying that much force upward would put the object in static equilibrium. The answer is, to lift the object upward you will need to apply a force greater than is

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