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Physics work and energy help?

(2) John slides a 10 kg object 5.0 meters horizontally across a level floor at a constant speed of 2 m/s using a force of 3.0 Newtons. How much work did John do on the object?

(3) Jill holds a 5 kg object 1.2 meters above the floor for 10 seconds, then slowly lowers it to the ground. How much work did Jill do on the object?

(4) A 20 Newton stone is pushed across level ground at a constant speed of 5 m/s. The pushing force required to do this is 40 Newtons. What was due net work done on the stone if it moves a distance of 10 meters?

(5) A 600 N force, directed parallel to the slope, is used to push an 80 kg box up an incline at constant speed. Approximately how much work was done against friction when moving the box from the bottom to the top of the incline?

(6) Consider an incline like the one in question 5 above, but frictionless, Also suppose the 80 kg box is at the top of the incline and someone pushes it down the hill with an initial speed of 4m/s. What is box's speed at the bottom of the incline?

2 Answers

• (2) 15 J
(3) 5*9.81*1.2 = 58.86 J
(4) 400 J
(5) data insufficient
(6) data insufficient, length or height of the incline needs to be given
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• (2) John slides a 10 kg object 5.0 meters horizontally across a level floor at a constant speed of 2 m/s using a force of 3.0 Newtons. How much work did John do on the object?
work W = F*d = 3*5 = 15 joule
since Ek = m/2*V^2 = 10/2*2^2 = 20 joule > 15 joule (W) it means the the mass was moving already @ 2m/sec

(3) Jill holds a 5 kg object 1.2 meters above the floor for 10 seconds, then slowly lowers it to the ground. How much work did Jill do on the object?
while hoisting : m*g*h = 6*10 = 60 joule
while lowering :-m*g*h = -6*10 = -60 joule
total work balance = 0

(4) A 20 Newton stone is pushed across level ground at a constant speed of 5 m/s. The pushing force required to do this is 40 Newtons. What was due net work done on the stone if it moves a distance of 10 meters?
work W = F*d = 40*10 = 400 joule

(5) A 600 N force, directed parallel to the slope, is used to push an 80 kg box up an incline at constant speed. Approximately how much work was done against friction when moving the box from the bottom to the top of the incline?
work = (F-m*g*sin Θ)*10

(6) Consider an incline like the one in question 5 above, but frictionless, Also suppose the 80 kg box is at the top of the incline and someone pushes it down the hill with an initial speed of 4m/s. What is box's speed at the bottom of the incline?
Vfin = √Vin^2+2gh = √4^2+20*h
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