A heavy-duty stapling gun uses a 0.162-kg metal rod that rams against the staple to eject it. The rod is pushed by a stiff spring called a "ram spring" (k = 32657 N/m). The mass of this spring may be ignored. Squeezing the handle of the gun first compresses the ram spring by 3.0 10-2m from its unstrained length and then releases it. Assuming that the ram spring is oriented vertically and is still compressed by 1.3 10-2m when the downward-moving ram hits the staple, find the speed of the ram at the instant of contact.
use this equation
v = x square root(k/m)
and the first x
v=?
x=3.0 x 10-2 m
k=32657 N/m
m=0.162 kg
v = x square root(k/m)
= (0.030m) x ((square root of)32657/0.162
=(0.030m)(square root of 201586.4198)
=(0.030m)(448.9837634)
=13.4695129 m/s
= 13 m/s
Why did the stapling gun go to therapy?
Because it had some serious spring compression issues!
To find the speed of the ram at the instant of contact, we can use the principle of conservation of mechanical energy.
The mechanical energy of the system is conserved because there are no external forces doing work on the system (ignoring air resistance and friction):
Initial mechanical energy (when the spring is compressed) = Final mechanical energy (when the ram hits the staple)
Initial mechanical energy (Ei) includes the potential energy stored in the compressed spring:
Ei = 1/2 k x^2
where k is the spring constant and x is the compression distance.
Final mechanical energy (Ef) includes the kinetic energy of the ram when it hits the staple plus the potential energy stored in the compressed spring (which is still partially compressed):
Ef = 1/2 mv^2 + 1/2 k x^2
where m is the mass of the metal rod and v is its velocity.
Since the potential energy stored in the spring is equal to the work done to compress it, we can set Ei equal to Ef:
1/2 k x^2 = 1/2 mv^2 + 1/2 k x^2
The x^2 terms cancel out:
mv^2 = 0
This implies that the velocity v is zero, which means that the ram comes to rest at the instant of contact with the staple.
Therefore, the speed of the ram at the instant of contact is zero.
To find the speed of the ram at the instant of contact, we can use the principle of conservation of mechanical energy. The initial potential energy stored in the compressed ram spring will be converted to the kinetic energy of the ram as it moves downwards and hits the staple.
First, let's calculate the potential energy stored in the ram spring when it is compressed by 3.0 x 10^-2 m. The formula for potential energy in a spring is:
Potential Energy (PE) = (1/2) k x^2
Where k is the spring constant and x is the displacement from the equilibrium position.
PE = (1/2) k x^2
= (1/2) (32657 N/m) (0.03 m)^2
= 14.7851 J (approximately)
Since the spring is still compressed by 1.3 x 10^-2 m when the ram hits the staple, this means that a portion of the potential energy is still stored in the spring. To find the kinetic energy of the ram at the instant of contact, we need to subtract this remaining potential energy from the initial potential energy.
Remaining Potential Energy = (1/2) k x^2
= (1/2) (32657 N/m) (0.013 m)^2
= 8.8724545 J (approximately)
Kinetic Energy (KE) = Initial Potential Energy - Remaining Potential Energy
= 14.7851 J - 8.8724545 J
= 5.9126455 J (approximately)
Now, we can use the formula for kinetic energy to calculate the speed of the ram at the instant of contact. The formula is:
KE = (1/2) m v^2
Where m is the mass of the ram and v is the speed.
5.9126455 J = (1/2) (0.162 kg) v^2
Simplifying the equation:
2(5.9126455 J) = (0.162 kg) v^2
11.825291 J = (0.162 kg) v^2
v^2 = 11.825291 J / 0.162 kg
v^2 ≈ 72.823153
v ≈ √(72.823153)
v ≈ 8.534 m/s
Therefore, the speed of the ram at the instant of contact is approximately 8.534 m/s.