Conservation of Energy, Work done on a System by an External Force

[Snomann92]

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°)
E_i=E_p
E_f=E_p'+E_e'
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?

E_i=1/2*k*x²
E_f=1/2*m*v²
1/2*m*v² = 1/2*k*x²
v=x*\sqrt{k/m}
v=(0.011m)*\sqrt{k/m} and v'=x'*\sqrt{k/m}

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'*\sqrt{k/m}/(0.011m)*\sqrt{k/m}=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|
F_a = \mu_k(F_g + F_a,y)
F_a = \mu_k*m*g + \mu_k*F_a*sin(28°)
F_a*(1 - \mu_k*sin(28°))= \mu_k*m*g
F_a = (\mu_k*m*g) / (1 - \mu_k*sin(28°))
F_a = ~58.40378N
W = F^a*d*cos(28°)
W = 382J

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All three came back wrong. :/

 

[D H]

All three came back wrong (I assume you have some automated system into which you are entering your answers) because as far as I can tell, you do have all three wrong. It would help if you attached the figures.

For the first problem, draw a right triangle that has the distance the box travels and the distance the box falls as two sides. Which of those is the hypotenuse and which is a leg of the triangle?

For the second problem, you have over-corrected the first attempted shot. You double-booked the correction. Your compression will result in missing the box by 22 cm, but too long rather than the first shot which was 22 cm short.

For the third problem, you did something very wrong from the very start. It appears that you are double-booking the coefficient of friction and it also appears you have some trigonometry errors. Without diagrams and without a description of your terms (what is F_a?) it is a bit hard to tell. Set up your solution a bit more carefully.

 

[Snomann92]

Thanks. 4am probably wasn't the best time to be doing physics...