A 2.0 kg block on a horizontal frictionless surface is attached to a spring whose force constant is 590 N/m. The block is pulled
1 answer:
Answer:
The value is
Explanation:
From the question we are told that
The mass of the block is
The force constant of the spring is
The amplitude is
The time consider is
Generally the angular velocity of this block is mathematically represented as
=>
=>
Given that the block undergoes simple harmonic motion the velocity is mathematically represented as
=>
=>
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Remember your kinematic equations for constant acceleration. One of the equations is 
, where 
= final position, 
= initial position, 
= initial velocity, t = time, and a = acceleration.
Your initial position is where you initially were before you braked. That means = 100m. You final position is where you ended up after t seconds passed, so = 350m. The time it took you to go from 100m to 350m was t = 8.3s. You initial velocity at the initial position before you braked was = 60.0 m/s. Knowing these values, plug them into the equation and solve for a, your acceleration:
Your acceleration is approximately
.
Your initial position is where you initially were before you braked. That means
Your acceleration is approximately
Answer:
a = 0.16
Explanation:
given,
mass of the object 1 = 0.2 kg
mass of the object 2 = 0.3 kg
acceleration when force is on 0.2 kg = 0.4 m/s²
acceleration when both mass are combine = ?
F = m a
F = 0.2 × 0.4
F = 0.08 N
force acting is same and total mass = 0.2 + 0.3 = 0.5 Kg
F = m a
a = 0.16 m/s²
the acceleration acting when both the body is attached is a = 0.16