A 2.5-kg, frictionless block is attached to an ideal spring with force constant 315N/m is undergoing simple harmonic motion. When the block has displacement 0.27 m, it is moving in the negative x-direction with a speed 4 m/s
part a: find the amplitude of the motion ? (........m)
part b: find the magnitude of the maximum force the spring exerts on the block? (..........N)
A 2.5-kg, frictionless block is attached to an ideal spring with force constant 315N/m is undergoing...
A 2.5-kg, frictionless block is attached to an ideal spring with force constant 315N/m is undergoing simple harmonic motion. When the block has displacement 0.27 m, it is moving in the negative x-direction with a speed 4 m/s part a: find the amplitude of the motion ? (........m) part b: find the magnitude of the maximum force the spring exerts on the block? (..........N) (I have only 1 left try in mastering physics, please help me thanks)
A 2.40 kg frictionless block is attached to an ideal spring with force constant 317 N/m . Initially the block has velocity -3.61 m/s and displacement 0.210 m . Part A Find the amplitude of the motion. Part B Find the maximum acceleration of the block. Part C Find the maximum force the spring exerts on the block.
A 2.50 kg frictionless block is attached to an ideal spring with force constant 312 N/m . Initially the block has velocity -3.67 m/s and displacement 0.290 m . Find the amplitude of the motion. Find the maximum acceleration of the block. Find the maximum force the spring exerts on the block.
A 2.10-kg frictionless block is attached to an ideal spring with force constant 325 N/m. Initially the spring is neither stretched nor compressed, but the block is moving in the negative direction at 13.5 m/s. A. Find the amplitude of the motion. Express your answer in meters. B. Find the maximum acceleration of the block. Express your answer in meters per second squared. C. Find the maximum force the spring exerts on the block. Express your answer in newtons.
A 2.20 kg frictionless block is attached to an ideal spring with force constant 316 N/m . Initially the block has velocity -3.80 m/s and displacement 0.240 m . A. Find the amplitude of the motion. B. Find the maximum acceleration of the block. C. Find the maximum force the spring exerts on the block.
A block of mass 1.20 kg is attached to a horizontal spring that has force constant k = 300 N/m. The block moves on a horizontal frictionless surface. The maximum speed of the block during its motion is 5 m/s. What is the amplitude A of the simple harmonic motion of the block?
A frictionless block of mass 2.30 kg is attached to an ideal spring with force constant 300 N/m . At t=0the spring is neither stretched nor compressed and the block is moving in the negative direction at a speed of 12.1 m/s . A. Find the amplitude. A =____ m B. Find the phase angle. ϕ = ____ rad C. Multiple Choice: Write an equation for the position as a function of time. (a.) x=(− 1.06 m )sin(( 11.4 rad/s...
A 2.00-kg, frictionless block is attached to an ideal spring with force constant 300 . At the block has velocity -4.00 and displacement +0.200 .Part AFind (a) the amplitude and (b) the phase angle.=Part Bφ=Part CWrite an equation for the position as a function of time.Assume in meters and in seconds.=
A block attached to an ideal spring of force constant (spring constant) 15 N/m executes simple harmonic motion on a frictionless horizontal surface. At time t = 0 s, the block has a displacement of -0.90 m, a velocity of -0.80 m/s, and an acceleration of +2.9 m/s2 . The mass of the block is closest to? Please show all of your work step by step including formulas used and variables used. A) 2.3 kg B) 2.6 kg C) 4.7...
A 2.00-kg, frictionless block is attached to an ideal spring with force constant 300 N/m. At t=0 the block has velocity -4.00 m/s and displacement +0.200 m. Part A Find (a) the amplitude and (b) the phase angle. A A = nothing m SubmitRequest Answer Part B ϕ ϕ = nothing rad SubmitRequest Answer Part C Write an equation for the position as a function of time. Assume x(t) in meters and t in seconds. x(t) x(t) = nothing m