The figure shows an 7.7 kg stone at rest on a spring. The spring is compressed 1

Question

The figure shows an 7.7 kg stone at rest on a spring. The spring is compressed 12 cm by the stone. (a) What is the spring constant? (b) The stone is pushed down an additional 31 cm and released.What is the elastic potential energy of the compressed spring just before that release? (c) What is the change in the gravitational potential energy of the stone The figure shows an 7.7 kg stone at rest on a spring. The spring is compressed 12 cm by the stone. (a) What is the spring constant? (b) The stone is pushed down an additional 31 cm and released.What is the elastic potential energy of the compressed spring just before that release? (c) What is the change in the gravitational potential energy of the stone

Explanation / Answer

a) The spring constant is easy. It is measured in Newtons per centimeter.
8kg*9.8m/s^2=78.4N
78.4N/8cm=9.8N/cm

b) The elastic potential energy of a spring is defined as the integral of the force put into the spring over the distance compressed.
In other words, the work you put into the spring to compress it.
PE=.5*kx^2
PE=.5*9.8*30^2 = 441J

c) When the stone is released from the compressed state it has 441 Joules of potential energy and 0 Joules of Kinetic Energy. When it passes through the relaxed position of the spring, there is no spring potential energy. All of it has been converted to Kinetic Energy. The energy is then converted from KE to gravitational potential energy. At the maximum height, the stone stops moving, i.e. KE=0. Conserving energy, the gravitational potential energy is now...441Joules!

d) Gravitational Potential energy is also the amount of work done to move an object against the gravitational field. Again, that is force * distance.
PE= mgh
441=8*9.8*h
h= 5.6 m

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