The jumping frog toy, shown airborne in the figure, consists of a spring with on

Question

The jumping frog toy, shown airborne in the figure, consists of a spring with one end attached to a flat base and the other end attached to a suction cup at the bottom of the upper body designed to look like a frog. Pressing the upper body of the toy so that the suction cup attaches to the base stores potential energy in the spring. With the spring compressed, if the toy is placed on a flat surface, the suction cup eventually detaches from the base sending the a) toy jumping into the air The jumping from toy is an example of the conversion of potential energy, stored in the spring, into kinetic energy as the frog leaves the surface, into gravitational potential energy as the frog toy reaches its peak height off the ground. We will estimate how far above the surface the frog will jump: (a) When the jumping frog toy's spring is separated from the base and upper body and attached firmly at one end, it stretches an additional 1.8 cm when a 500 g weight is hung from the other end. What is the spring constant, k? (b) The toy's spring is compressed relative to its fully expanded state by about 1 cm, when the suction cup detaches from the base. How much energy in millijoules is (c) If all of the potential energy in the spring when it detaches from the base is converted into the frog's potential energy at the highest point of the toy's trajectory, how high in centimeters will the frog jump? Note that the toy's mass in panel (b) of the figure is in grams.

Explanation / Answer

Solution) We have F=Kx

But F=mg so substituting F=mg we get

mg=Kx

Here m is mass

g is acceleration due to gravity

K is spring constant

x is distance

Substituting the values we get

K=mg/x

K=(500×10^(-3)kg)×9.8/(1.8×10^(-2))

K=272.22N/m

(b) Here energy is spring potential energy(PE)

PE=(1/2)Kx^2=(1/2)(272.22)(1×10^(-2))^2=13.611mJ

If changes from 1.8 cm to 1cm

PE=(1/2)K((1.8×10^(-2))^2 - (1×10^(-2))^2)=30.488mJ

(c)we have spring potential energy equal to frogs potential energy

(1/2)Kx^(2)=mgh

First case potential energy of spring=13.611mJ

13.611×10^(-3)=(500×10^(-3)×10×h)

h=0.0027m

Second case if potential energy of spring is 30.488mJ

30.488×10^(-3)=(500×10^(-3)×10×h)

h=0.006m

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