Use the exact values you enter in previous answer(s) to make later calculation(s

Question

Use the exact values you enter in previous answer(s) to make later calculation(s).

A toy gun shoots spherical plastic projectiles by means of a spring. A typical projectile has mass

m = 20 g.

The spring used has spring constant

k = 30 N/m,

and when put under load, it is displaced an amount

?x = 6.0 cm

as shown in the figure below. The barrel of the gun exerts a slight frictional force of magnitude

Ffriction = 0.072 N

on the pellet as it moves down the barrel from its starting point a total length

L = 15 cm.

(a) If the toy gun is fired in a horizontal position, at what position measured from the starting point does the pellet reach maximum velocity? Hint: Consider an equilibrium condition between the spring force and the friction force on the pellet before it leaves the spring.
cm

(b) What is the maximum speed achieved by the pellet as it is fired from the gun?
m/s

(c) At what speed does the pellet leave the barrel?
m/s

Explanation / Answer

mass=0.02 kg
k=30
delta_x=0.06 m
force of friction=0.072 N

as the spring is pulling the pellet ahead, the force on the pellet =30*(0.06-x) if pellet is moved by distance x from the initial position
at the same time an opposing frictional force 0.072 N will be acting
so initially the spring force will be higher and as it pushes the pellet forward. the spring force will start decreasing but as long as it is higher than the friction force, there will be positive accelration and speed will go on increasing
but when the spring force is equal to friction force and subsequently decreases and net force becomes negative , there will be deceleration and speed will decrease.
hence maximum speed will be achieved when spring force=friction force
then 30*(0.06-x)=0.072
x=5.76 cm
b)
as the pellet moves 5.76 cm, the energy imparted by the spring=0.5*30*(5.76*0.01)^2=49.766 mJ
work done against force of friction=0.072*5.76*0.01= 0.0041472 J
net kinetic energy of the pellet= 0.045619 J
so if speed is v,
then 0.5*0.02*v^2= 0.045619
v=2.136 m/s

c) now potential energy of the spring=0.5*k*delta_x^2=0.5*30*0.06^2=0.054 J

work done against friction force as the pellet covers the distance L=0.072*L=0.0108 J

so energy left with pellet when it is fired=0.054-0.0108=0.0432 J

then if speed is v,
then 0.5*0.02*v^2=0.0432

v=2.0785 m/s

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