Physics, 13.10.2020 03:01 wrightlilybug07

One end of a spring with a force constant of k = 10.0 N/m is attached to the end of a long horizontal frictionless track and the other end is attached to a mass m = 2.20 kg which glides along the track. After you establish the equilibrium position of the mass-spring system, you move the mass in the negative direction (to the left), compressing the spring 2.48 m. You then release the mass from rest and start your stopwatch, that is x(t = 0) = −A, and the mass executes simple harmonic motion about the equilibrium position. Determine the following.(a) displacement of the mass (magnitude and direction) 1.0s after it is released (b) velocity of the mass (magnitude and direction) 1.0s after it is released
(c) acceleration of the mass (magnitude and direction) 1.0s after it is released
(d) force the spring exerts on the mass (magnitude and direction) 1.0s after it is released
(e) How many times does the object oscillate in 12.0s?

Answers: 2

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Answers

Answer from: allytrujillo20oy0dib

-2.478

0.379

11.14

24.78

Explanation:

Angular frequency of spring in harmonic motion is given by?

ω = √(k/m)

ω = √(10/2.2)

ω = √4.54

ω = 2.13 s^-1

If at t=0 the mass is in negative amplitude (x = -A = -2.48 m) then we describe the position with negative cosine

x(t) = -A * cos(ωt)

x(t) = -2.48 * cos(2.13 * 1)

x(t) = -2.48 * 0.9993

x(t) = -2.478

Velocity and acceleration are 1st and 2nd derivative of position

v(t) = Aω * sin(ωt)

v(t) = 2.48 * 2.13 * sin(2.13 * 1)

v(t) = 5.282 * sin2.13

v(t) = 5.282 * 0.03717

v(t) = 0.379 m/s

a(t) = Aω^2 * cos(ωt)

a(t) = 2.48 * 2.12² * cos(2.13 * 1)

a(t) = 2.48 * 4.494 * cos2.13

a(t) = 11.15 * 0.9993

a(t) = 11.14 m/s²

F = -k * x(t)

F = -10 * -2.478

F = 24.78 N

Answer from: Quest

c) power lifting, will be your answer.

deku ❤

Answer from: Quest

m₁v₁ + m₂v₂ = (m₁ + m₂)vmomentum is conserved.the left-hand side of the equation represents the momentum in the system before the collision.the right-hand side represents the momentum in the system after the collisionm₁ and v₁ represents the mass and the velocity of the physics blockm₂ and v₂ represents the mass and velocity of the ball of puttyv represents the final velocity of the combined mass.the question states the two masses stick together. they combined mass has its own velocity with the body mass the sum of the two objects' massestheir combined momentum is (m₁ + m₂)vto find the magitude of (m₁ + m₂)v, we need to add the vectors on the left-hand side of the equation: m₁v₁ + m₂v₂.note the diagram. i have positioned the vectors in accordance with their directions.the magnitude of the resultant can be calculated using the pythagorean theorem since this is a right triangle situation (north and west are perpendicular)|(m₁ + m₂)v| = √[ (m₁v₁)² + (m₂v₂)² ]|(m₁ + m₂)v| = √[ (1.5kg·8.0m/s)² + (0.40kg·20m/s)² ]|(m₁ + m₂)v| = 14.4 kg m/sfor two significant figures (assuming that 20m/s is two), this rounds to 14 kg m/s