A spring with an 8-kg mass is kept stretched 0.4 m beyond its natural length by a force of 32 N. The spring starts at its equili
2 answers:
Answer:
x = 1.26 sin 3.16 t
Explanation:
Assume that the general equation of the displacement given as
x = A sinω t
A=Amplitude ,t=time ,ω=natural frequency
We know that speed V
V= A ω cosωt
Maximum velocity
V(max)= Aω
Given that F= 32 N
F = K Δ
K=Spring constant
Δ = 0.4 m
32 =0.4 K
K = 80 N/m
We know that ω²m = K
8 ω² = 80
ω = 3.16 s⁻¹
Given that V(max)= Aω = 4 m/s
3.16 A = 4
A= 1.26 m
Therefore the general equation of displacement
x = 1.26 sin 3.16 t
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Answer:
a. ,
b.
Explanation:
To calculate the values in the question, a deep understanding of perfect inelastic collision is important.
When two bodies undergo inelastic collision, two important parameters must be well understood i.e
Momentum: the momentum is always conserved in perfectly inelastic collision. i.e the total momentum after collision is the sum of the individual momentum before collision
Kinetic energy: Kinetic energy is not conserved due to dissipative force.
a.To calculate the velocity, we first find the total momentum before collision
Momentum of player 1
Momentum of player 2
Hence the total momentum
Note, since the direction of movement before collision is due south and due north respectively we have to represent the velocity using the rectangular coordinate
Hence
solving for the resultant velocity, we have
To calculate the direction of movement, we have
b. to calculate the decrease in total kinetic energy, before collision, the total kinetic was
And the final kinetic energy after collision is
The decrease in Kinetic energy is
The negative sign indicate a decrease in Kinetic energy