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Snomann92
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A 1.6 kg breadbox on a frictionless incline of angle θ = 36° is connected, by a cord that runs over a pulley, to a light spring of spring constant k = 120 N/m, as shown in the figure below. The box is released from rest when the spring is unstrectched. Assume that the pulley is massless and frictionless.
(b) How far down the incline from its point of release does the box slide before momentarily stopping?
h=x/sin(36°)
Ei=Ep
Ef=Ep'+Ee'
Therefore, m*g*h' + 1/2*k*x'2 = m*g*h
Since h=x/sin(36°) and x=0
m*g*x'/sin(36°) + 1/2*k*x'2 = 0
x'=-(m*g)/(k*sin(36°))
x'=0.44m
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Two children are playing a game in which they try to hit a small box on the floor with a marble fired from a spring-loaded gun that is mounted on a table. The target box is horizontal distance D = 2.00 m from the edge of the table, see the figure. Bobby compresses the spring 1.10 cm, but the center of the marble falls 22.0 cm short of the center of the box. How far should Rhoda compress the spring to score a direct hit?
Ei=1/2*k*x2
Ef=1/2*m*v2
1/2*m*v2 = 1/2*k*x2
v=x*[tex]\sqrt{k/m}[/tex]
v=(0.011m)*[tex]\sqrt{k/m}[/tex] and v'=x'*[tex]\sqrt{k/m}[/tex]
Since h = 1/2*g*t2 = 1/2*g*t'2,
t = t'
v'*t'/v*t = v'*t/V*t = v'/v = 2.20m/1.78m = ~1.2359
v'/v = x'*[tex]\sqrt{k/m}[/tex]/(0.011m)*[tex]\sqrt{k/m}[/tex]=x'/0.011m
Therefore, x' = 0.011m * ~1.2359
x' = 1.35949cm
x' = 1.36cm
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A worker pushed a 27 kg block 7.4 m along a level floor at constant speed with a force directed 28° below the horizontal.
(a) If the coefficient of kinetic friction is 0.20, how much work was done by the worker's force?
|applied force| = - |frictional force|
Fa = [tex]\mu[/tex]k(Fg + Fa,y)
Fa = [tex]\mu[/tex]k*m*g + [tex]\mu[/tex]k*Fa*sin(28°)
Fa*(1 - [tex]\mu[/tex]k*sin(28°))= [tex]\mu[/tex]k*m*g
Fa = ([tex]\mu[/tex]k*m*g) / (1 - [tex]\mu[/tex]k*sin(28°))
Fa = ~58.40378N
W = Fa*d*cos(28°)
W = 382J
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All three came back wrong. :/
(b) How far down the incline from its point of release does the box slide before momentarily stopping?
h=x/sin(36°)
Ei=Ep
Ef=Ep'+Ee'
Therefore, m*g*h' + 1/2*k*x'2 = m*g*h
Since h=x/sin(36°) and x=0
m*g*x'/sin(36°) + 1/2*k*x'2 = 0
x'=-(m*g)/(k*sin(36°))
x'=0.44m
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Two children are playing a game in which they try to hit a small box on the floor with a marble fired from a spring-loaded gun that is mounted on a table. The target box is horizontal distance D = 2.00 m from the edge of the table, see the figure. Bobby compresses the spring 1.10 cm, but the center of the marble falls 22.0 cm short of the center of the box. How far should Rhoda compress the spring to score a direct hit?
Ei=1/2*k*x2
Ef=1/2*m*v2
1/2*m*v2 = 1/2*k*x2
v=x*[tex]\sqrt{k/m}[/tex]
v=(0.011m)*[tex]\sqrt{k/m}[/tex] and v'=x'*[tex]\sqrt{k/m}[/tex]
Since h = 1/2*g*t2 = 1/2*g*t'2,
t = t'
v'*t'/v*t = v'*t/V*t = v'/v = 2.20m/1.78m = ~1.2359
v'/v = x'*[tex]\sqrt{k/m}[/tex]/(0.011m)*[tex]\sqrt{k/m}[/tex]=x'/0.011m
Therefore, x' = 0.011m * ~1.2359
x' = 1.35949cm
x' = 1.36cm
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A worker pushed a 27 kg block 7.4 m along a level floor at constant speed with a force directed 28° below the horizontal.
(a) If the coefficient of kinetic friction is 0.20, how much work was done by the worker's force?
|applied force| = - |frictional force|
Fa = [tex]\mu[/tex]k(Fg + Fa,y)
Fa = [tex]\mu[/tex]k*m*g + [tex]\mu[/tex]k*Fa*sin(28°)
Fa*(1 - [tex]\mu[/tex]k*sin(28°))= [tex]\mu[/tex]k*m*g
Fa = ([tex]\mu[/tex]k*m*g) / (1 - [tex]\mu[/tex]k*sin(28°))
Fa = ~58.40378N
W = Fa*d*cos(28°)
W = 382J
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All three came back wrong. :/