a 4kg roller is attached to a spring of force constant 100N/m. it slides over a frictionless horizontal road. the roller is displaced from equilibrium position by 10cm and then released. its maximum speed will be?
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A=0.1 m
ω=sqrt(k/m) =sqrt(100/4) =5 rad/s
v(max) =Aω =0.1•5=0.5 m/s
To find the maximum speed of the roller, we need to consider the potential energy stored in the spring and the conservation of mechanical energy.
The potential energy stored in a spring is given by the formula:
PE = (1/2) k x^2
Where:
PE is the potential energy
k is the force constant of the spring
x is the displacement from the equilibrium position
In this case, the displacement from the equilibrium position is 10 cm, which is equivalent to 0.1 meters.
So, the potential energy stored in the spring is:
PE = (1/2) * 100 N/m * (0.1 m)^2
= (1/2) * 100 N/m * 0.01 m^2
= 0.5 J
The mechanical energy of the roller is conserved, which means that the potential energy is converted to kinetic energy when the roller is released.
The maximum speed of the roller can be found by equating the potential energy to the kinetic energy:
KE = (1/2) m v^2
Where:
KE is the kinetic energy
m is the mass of the roller
v is the velocity of the roller
In this case, the mass of the roller is 4 kg.
Equating the potential energy to the kinetic energy, we have:
0.5 J = (1/2) * 4 kg * v^2
Simplifying the equation, we get:
v^2 = 0.5 J / (2 kg)
v^2 = 0.25 J/kg
Taking the square root of both sides of the equation, we find:
v = √( 0.25 J/kg )
v ≈ 0.5 m/s
Therefore, the maximum speed of the roller will be approximately 0.5 m/s.