In the figure shown below, a 520 g box is placed in front of a spring, with spri

Question

In the figure shown below, a 520 g box is placed in front of a spring, with spring constant 5.5 N/cm, that is compressed a distance delta x from its natural length. When the box is let go, the spring will push the box and it will slide up the frictionless hill of height H = 1.5 m. At the top of the hill, the ground is not frictionless. The coefficient of kinetic friction in this area is 0.15 and the coefficient of static friction is 0.25. The box will travel a distance d = 7.0 cm on the rough surface at the top of the hill before coming to a stop. What is the total work that friction does in bringing the box to a stop? W = 1/2 Kx^2 = -1/2 (5.5) (59)^2 Before entering the area with friction what is the kinetic energy of the box? KE = 1/2 mv^2 = 1/2 (520)(5.5)^2 = 7865 J What is the total mechanical (potential plus kinetic) energy of the box at the top of the hill? the kinetic energy is the same as potential enegy = -7865 S What is the total mechnical energy at the bottom of the hill (after leaving the spring)? 1 + is zero. How much potential energy is stored in the spring when it is fully compressed? What distance, delta x is the spring intially compressed? U = 1/2 kd^2 = 1/2 (55)(7.0)^2 = 134.75 = U - ma = 134.75 - (0.520)(9.8) = 130 = 130/0.15 (9.8)(1.5) = 59

Explanation / Answer

a) work done by friction force= umg*d=0.15*0.52*9.8*0.07= 0.053 J
b)kinetic energy = work done by friction force= umg*d=0.15*0.52*9.8*0.07= 0.053 J
c)mechanical energy=kinetic energy+ potential energy=0.053+0.52*9.8*1.5= 7.697 J(before
passing friction full surface)
d)mechanical energy at bottom = 7.697 J (energy is conserved)
e)potential energy = 7.697 J(since there is no kinetic energy)
f)(1/2)*k*x^2=7.697
x= 0.0529 m

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