The figure shows an 7.8 kg stone at rest on a spring. The spring is compressed 9

Question

The figure shows an 7.8 kg stone at rest on a spring. The spring is compressed 9.6 cm by the stone. (a) What is the spring constant? (b) The stone is pushed down an additional 27 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–Earth system when the stone moves from the release point to its maximum height? (d) What is that maximum height, measured from the release point?

Explanation / Answer

Answer:-

Given Data:-

m = 7.8 kg

x = -9.6 cm = -0.096 m

(a) What is the spring constant?

we know the formula F = k*x

F = m*g = 7.8*9.8 = 76.44 N

Hence so spring constant k = F/x = -76.44/ -0.096

k = 796.25 N/m

(b) The stone is pushed down an additional 27 cm and released.What is the elastic potential energy of the compressed spring just before that release?

here we have to calculate the potential energy when stone pushed down.

so we will consider the x value x= 27 cm = 0.27 m

we know a potential energy formula for spring. P.E = 1/2*k*x2

P.E = 0.5* 796.25*0.272

P.E = 29.02 J.

(c) What is the change in the gravitational potential energy of the stone–Earth system when the stone moves from the release point to its maximum height?

A stone has the potential energy 29.2 J whe it is in compressed state at this time it has no kinetic energy means zero J of kinetic energy. so at the maximum stone has relaxed and stone stops moving with zero kinetic energy.

So we can state that the gravitational potential energy would be 29.2 J.

(d) What is that maximum height, measured from the release point?

In part-c we have consider the gravitational potential energy at the maximum height is 29.3 .

now we have to calculate the maximum height so conside the formula of potential energy due to gravity

P.E = m*g*h it is consdier as the ( F*d)

put he P.E value of part-c here.

29.2 = 7.8*9.8 *h

h =0.38 m

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