A 2.60-kg block is attached to a horizontal spring that has a spring constant of 160. N/m. At the instant when the displacement of the spring from its unstrained length is -0.115 m, what is the acceleration a of the object?
force on the spring:kx
but force= mass*acceleration
so acceleration=kx/mass
check my thinking.
To find the acceleration of the object, we need to 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 formula for Hooke's Law is:
F = -kx
Where:
- F is the force exerted by the spring,
- k is the spring constant, and
- x is the displacement of the spring from its equilibrium position.
In this case, the displacement of the spring is given as -0.115 m, and the spring constant is 160 N/m.
Plugging these values into the equation, we have:
F = -(160 N/m)(-0.115 m)
F = 18.4 N
Since the force exerted by the spring is equal to the mass multiplied by the acceleration (F = ma), we can rearrange the equation to solve for acceleration:
a = F / m
Given that the mass of the block is 2.60 kg, we have:
a = (18.4 N) / (2.60 kg)
a = 7.077 m/s^2
Therefore, the acceleration of the object is 7.077 m/s^2.