The acceleration of the particle at time t is:

The velocity of the particle at time t is given by the integral of the acceleration a(t):

and the position of the particle at time t is given by the integral of the velocity v(t):

Assuming the particle starts from position x(0)=0 at t=0, the distance the particle covers in the first t=2 seconds can be found by substituting t=2 s in the equation of x(t):
Answer:
Explanation:
mass m = 3 kg
spring constant be k
k x .8 = 40 N
k = 40 / .8 = 50 N /m
angular frequency ω = √ ( k / m )
= √ ( 50 / 3 )
= 4.08 rad /s
Let amplitude of oscillation be A .
1/2 k A² = 1/2 m v²
50 A² = 3 x 1²
A = .245 m = 24.5 cm
For displacement , the equation of SHM is
x = A sinωt
= 24.5 sin4.08 t
x = 24.5 sin4.08 t
Here, angle 4.08 t is in radians .
Answer:
0.3 m
Explanation:
Initially, the package has both gravitational potential energy and kinetic energy. The spring has elastic energy. After the package is brought to rest, all the energy is stored in the spring.
Initial energy = final energy
mgh + ½ mv² + ½ kx₁² = ½ kx₂²
Given:
m = 50 kg
g = 9.8 m/s²
h = 8 sin 20º m
v = 2 m/s
k = 30000 N/m
x₁ = 0.05 m
(50)(9.8)(8 sin 20) + ½ (50)(2)² + ½ (30000)(0.05)² = ½ (30000)x₂²
x₂ ≈ 0.314 m
So the spring is compressed 0.314 m from it's natural length. However, we're asked to find the additional deformation from the original 50mm.
x₂ − x₁
0.314 m − 0.05 m
0.264 m
Rounding to 1 sig-fig, the spring is compressed an additional 0.3 meters.
The answer is 0.025J.
W=1/2*k*x^2
W=1/2*20*0.050^2
W=0.025J