A 3.7-kg sphere is suspended by a cord that passes over a 1.3-kg pulley of radius 3.6 cm. The cord is attached to a spring whose force constant is k = 86 N/m as in the figure below. Assume the pulley is a solid disk.An apparatus is shown with a wall extending upward from the left end of a table and a pulley mounted above the right edge of the table. A horizontal spring labeled k has its left end attached to a horizontal cord, the other end of which is attached to the wall. The center line of the spring is aligned with the top of the pulley. The right end of the spring is attached to a second cord. This cord extends horizontally, over the pulley and 90° around it where it hangs downward. A sphere of mass m hangs from the end of the second cord.(a) If the sphere is released from rest with the spring unstretched, what distance does the sphere fall through before stopping? m(b) Find the speed of the sphere after it has fallen 25 cm. m/s
Question
A 3.7-kg sphere is suspended by a cord that passes over a 1.3-kg pulley of radius 3.6 cm. The cord is attached to a spring whose force constant is k = 86 N/m as in the figure below. Assume the pulley is a solid disk.An apparatus is shown with a wall extending upward from the left end of a table and a pulley mounted above the right edge of the table. A horizontal spring labeled k has its left end attached to a horizontal cord, the other end of which is attached to the wall. The center line of the spring is aligned with the top of the pulley. The right end of the spring is attached to a second cord. This cord extends horizontally, over the pulley and 90° around it where it hangs downward. A sphere of mass m hangs from the end of the second cord.(a) If the sphere is released from rest with the spring unstretched, what distance does the sphere fall through before stopping? m(b) Find the speed of the sphere after it has fallen 25 cm. m/s
Solution
(a) To find the distance the sphere falls through before stopping, we need to equate the potential energy gained by the sphere to the potential energy stored in the spring when the sphere stops.
The potential energy gained by the sphere is given by mgh, where m is the mass of the sphere, g is the acceleration due to gravity, and h is the height the sphere falls.
The potential energy stored in the spring is given by (1/2)kx^2, where k is the spring constant and x is the distance the spring is stretched.
Setting these two equal to each other gives:
mgh = (1/2)kx^2
Solving for h (the distance the sphere falls) gives:
h = (1/2)kx^2 / (mg)
We know that when the sphere stops, the spring is stretched by the same distance the sphere falls, so x = h. Substituting this into the equation gives:
h = (1/2)k*h^2 / (mg)
Solving for h gives:
h = sqrt((2mg)/k)
Substituting the given values gives:
h = sqrt((23.79.8)/86) = 0.98 m
(b) To find the speed of the sphere after it has fallen 25 cm, we can use the conservation of energy. The potential energy lost by the sphere is equal to the kinetic energy gained by the sphere plus the potential energy stored in the spring.
The potential energy lost by the sphere is mgh, the kinetic energy gained by the sphere is (1/2)mv^2, and the potential energy stored in the spring is (1/2)kx^2.
Setting these equal to each other gives:
mgh = (1/2)mv^2 + (1/2)kx^2
Solving for v (the speed of the sphere) gives:
v = sqrt((2mgh - kx^2)/m)
Substituting the given values gives:
v = sqrt((23.79.80.25 - 860.25^2)/3.7) = 1.22 m/s
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