Solve the following problem: The launching mechanism of a toy gun consists of a

Question

Solve the following problem: The launching mechanism of a toy gun consists of a spring as shown in the figure. To load the gun you have to compress the spring a distance ofx 0.12 m by applying a force F 65N. Then the gun is fired vertically as shown launching a 20.0-g projectile from rest to a maximum height H above the starting point of the projectile. Determine the spring constant. Determine how much energy is stored in a spring when it is compressed by a distance x. Maks sure to include the appropriate units. a. b. c. Neglecting all resistive forces, describe the mechanical energy transformations that occur from the time the gun is fired until the projectile reaches its maximum height, H d. Neglecting all resistive forces, find the speed of the projectile as it moves through the equilibrium position of the spring (where x = 0), as shown in Figure (b). Neglecting all resistive forces, determine the maximum height H. e. Neglecting all rsistive forces, determine the maximum e.

Explanation / Answer

Use the law of conservation of energy, i.e.,

Energy from spring = energy absorbed by projectile

(1/2)kx^2 = mgh

where

k =spring constant
x = compressed length of spring = 0.12 m
m = mass of projectile = 20 g (given) = 0.020 kg.
g = acceleration due to gravity = 9.8 m/sec^2 (constant)
h = maximum height attained by projectile = H m.

Substituting values,

(1/2)(k)(0.12)^2 = (0.020)(9.8)(H)

k = 2*(0.020)(9.8)(H)/0.12^2

k = 27.22 H N/m

<< Neglecting all resistive forces, determine the speed of the projectile as it moves through the equilibrium position of the spring (where x = 0), as shown in figure (b). Assume for this question that the spring constant, k, = 27.22 H N/m. >>

Again, use the law of conservation of energy.

Energy from spring = Energy absorbed by projectile

(1/2)(k)(x^2) = (1/2)(m)(V^2)

V^2 = 27.22 H(0.12)/0.020

without H we can't solve it

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