i have a spring loaded toy gun that propels projectiles of mass .001 kg. the spr

Question

i have a spring loaded toy gun that propels projectiles of mass .001 kg. the spring has a spring constant of 10 N/m and when ready to fire the spring is compressed .1m. I fire the toy gun straight up into the air.

a. what is the change in potential energy of the projectile from the time just before i pull the trigger to when the projectile leaves the barrel? b. what is the speed of the projectile as it leaves the barrel? c. how high will my projectile go? d. when the projectile is at its highest, where did the kinetic energy that it had as it left the barrel go?

Explanation / Answer

a) change in potential energy of the projectile = m*g*x

= 0.001*9.8*0.1

= 9.8*10^-4 J

b)

Apply conservation of energy

initail mechaincal energy = final mechaincal energy

0.5*k*x^2 = m*g*x + 0.5*m*v^2

0.5*10*0.1^2 = 9.8*10^-4 + (1/2)*0.001*v^2

5*10^-2 = 9.8*10^-4 + 5*10^-4*v^2

v = sqrt( (5*10^-2 - 9.8*10^-4)/(5*10^-4) )

= 9.9 m/s

c) maximum height reached by the projectile, hmax = v^2/(2*g)

= 9.9^2/(2*9.8)

= 5 m

d) the initial kinetic energy is converted to final gravitational potential energy.

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