Initially sliding with a speed of 2.5 m/s, a 2.0 kg block collides with a spring and compresses it 0.35 m before coming to rest. What is the force constant of the spring?
K= N/M
Two equations:
KE=1/2mv^2
and
PE=1/2kx^2
Where:
PE=potential energy
KE=Kinetic energy
m=mass of the block
v=velocity of the block
k=spring constant
and
x=0m-0.35m=-0.35m
Set both equations equal together and solve for k:
1/2mv^2=-1/2kx^2
k=mv^2/x^2
k=[2.0kg*(2.5m/s)^2]/(0.35m)^2
Answer should contain no more than two significant figures.
To find the force constant of the spring, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position.
The equation for Hooke's Law is:
F = -kx
Where:
F = Force exerted by the spring
k = Force constant (also known as the spring constant)
x = Displacement of the spring
In this case, the block comes to rest after compressing the spring, so the final velocity is 0 m/s.
To find the force constant (k), we need to determine the force exerted by the spring, which can be calculated using Newton's second law:
F = ma
Where:
F = Force exerted by the spring
m = Mass of the object (2.0 kg in this case)
a = Acceleration of the object
Since the final velocity is 0 m/s, the acceleration is given by the equation:
v^2 = u^2 + 2as
Where:
v = Final velocity (0 m/s)
u = Initial velocity (2.5 m/s in this case)
a = Acceleration
s = Displacement (0.35 m in this case)
Rearranging the equation to solve for acceleration (a), we have:
0 = (2.5)^2 + 2a(0.35)
0 = 6.25 + 0.7a
0.7a = -6.25
a = -6.25 / 0.7
a ≈ -8.93 m/s^2
Now, we can substitute the mass (m) and acceleration (a) into Newton's second law to find the force exerted by the spring:
F = ma
F = 2.0 kg * (-8.93 m/s^2)
F ≈ -17.86 N
Since we are interested in the force exerted by the spring, which is negative due to the direction, we use the negative value.
Finally, we can substitute the force (F) and the displacement (x) into Hooke's Law to find the force constant (k):
-17.86 N = -k * 0.35 m
k = -17.86 N / 0.35 m
k ≈ 51.03 N/m
Therefore, the force constant of the spring is approximately 51.03 N/m.