1) toy gun fires a 120 g projectile disc by using a compressed spring (k= 1700N/m). The barrel of the gun is 5.0 cm. As the projectile goes down the barrel, it creates thermal energy due to friction. The frictional force is 0.13N. You compress the spring 84mm 1) What is the disc's speed when it leaves the barrel of the gun 2) What would be the disc's max height if shot straight up (assuming a vacuum)

See the explanation if you have any dought please ask in the comment section

Step-by-step explanation

Frictional force, f = 0.13 N

Length of the barrel, h = 5 cm =0.05 m

So when the projectile is at the bottom of the barrel the thermal energy stored is,

E_T = f*h = 0.13*0.05 J = 0.0065 J

..............................(1)

If the spring compress,

?x = 84 mm = 0.084 m

Energy stored in the spring gun is

E_S = (1/2)k*?x² = 0.5*1700*0.084² J

= 5.9976 J

So total energy, E = E_T + E_S = 6.0041J

Let the velocity of the projectile of mass, m = 120 g = 0.12 kg is V

So kinetic energy,

E_K = (1/2)mV²

Conservation of energy shows,

E_K = E

=> (1/2)mV² =6.0041

=> V = (2*6.0041/m)^0.5

= (2*6.0041/0.12)^0.5 m/s

= 10 m/s

b) since the projectile is shot straight up the maximum height is

H = V²/2g = 10²/2*9.8 m = 5.10 m