A mass of 6.10 kg rests on a smooth surface inclined 39.0o above the horizontal. It is kept from sliding down the plane by a spring attached to the wall. the spring is aligned with the plane and has a spring constant of 126 N/m. How much does the spring stretch?
can you show the steps to how to answer the question as well ? just the steps would be fine :)
vector m•a=vector m•g +vector F(spring)
a=0 =>
0=vector m•g +vector F(spring)
x: 0 = m•g•sinα –k•x,
x=m•g•sinα/k=6.1•9.8•sin 39°/126 =0.3 m
Fm = M*g = 6.10 * 9.8 = 59.8 N. = Force of the mass.
Fp = 59.8*sin39 = 37.63 N. = Force parallel to the incline = Force applied to the spring.
d=37.63/126 * 1m = 0.30 m. = Distance the spring is stretched.
To find out how much the spring stretches, we need to analyze the forces acting on the mass.
First, let's resolve the weight of the object into components. The weight can be divided into two components: one parallel to the inclined plane (mg sin θ) and the other perpendicular to the inclined plane (mg cos θ).
The force along the inclined plane (parallel to the plane) can be found using the equation:
F_parallel = mg sin θ
where m is the mass (6.10 kg) and θ is the angle of inclination (39.0 degrees). Plugging in the values:
F_parallel = (6.10 kg) * 9.8 m/s^2 * sin(39.0 degrees)
= 37.64 N
This force is balanced by the force exerted by the spring. According to Hooke's Law, the force exerted by a spring is proportional to the displacement from its equilibrium position. The equation can be written as:
F_spring = -k * x
where F_spring is the force exerted by the spring, k is the spring constant (126 N/m), and x is the displacement (stretching) of the spring.
Since the mass is not moving, the force exerted by the spring is equal in magnitude and opposite in direction to the force along the inclined plane:
F_spring = F_parallel
-k * x = mg * sin θ
Rearranging the equation, we can solve for x:
x = (-mg * sin θ) / k
Plugging in the values:
x = (-6.10 kg * 9.8 m/s^2 * sin(39.0 degrees)) / (126 N/m)
= -0.672 m
The negative sign in the answer indicates that the spring is compressed by 0.672 m.