Explanation:
(a) Hooke's law:
F = kx
7.50 N = k (0.0300 m)
k = 250 N/m
(b) Angular frequency:
ω = √(k/m)
ω = √((250 N/m) / (0.500 kg))
ω = 22.4 rad/s
Frequency:
f = ω / (2π)
f = 3.56 cycles/s
Period:
T = 1/f
T = 0.281 s
(c) EE = ½ kx²
EE = ½ (250 N/m) (0.0500 m)²
EE = 0.313 J
(d) A = 0.0500 m
(e) vmax = Aω
vmax = (0.0500 m) (22.4 rad/s)
vmax = 1.12 m/s
amax = Aω²
amax = (0.0500 m) (22.4 rad/s)²
amax = 25.0 m/s²
(f) x = A cos(ωt)
x = (0.0500 m) cos(22.4 rad/s × 0.500 s)
x = 0.00919 m
(g) v = dx/dt = -Aω sin(ωt)
v = -(0.0500 m) (22.4 rad/s) sin(22.4 rad/s × 0.500 s)
v = -1.10 m/s
a = dv/dt = -Aω² cos(ωt)
a = -(0.0500 m) (22.4 rad/s)² cos(22.4 rad/s × 0.500 s)
a = -4.59 m/s²
Answer:
6 m/s
Explanation:
Given that :
mass of the block m = 200.0 g = 200 × 10⁻³ kg
the horizontal spring constant k = 4500.0 N/m
position of the block (distance x) = 4.00 cm = 0.04 m
To determine the speed the block will be traveling when it leaves the spring; we applying the work done on the spring as it is stretched (or compressed) with the kinetic energy.
i.e 





v = 6 m/s
Hence,the speed the block will be traveling when it leaves the spring is 6 m/s
Answer:
1.11 m/s
Explanation:
The motion of the boat is an example of accelerated motion, since the velocity is not constant. However, we don't need to find the acceleration, because we are only interested in the average velocity of the boat, which is given by:

where d is the total distance covered and t the time taken. In this problem, the boat covered a distance of d = 20 m and it takes t = 18 s, therefore the average velocity is

Answer:
The momentum of block B = 27 Kg m/s
Explanation:
Given,
The initial momentum of block A, MU = 15 Kg m/s
The final momentum of block A, MV = -12 Kg m/s
Consider the block B is initially at rest.
Therefore, the initial momentum of block B, mu = 0
According to the laws of conservation of linear momentum, the momentum of the body before impact is equal to the momentum of the body after impact.
<em> MU + mu = MV + mv</em>
15 + (0) = (-12) + mv
mv = 15 + 12
= 27 Kg m/s
Hence, the momentum of the block B after impact is, mv = 27 Kg m/s
Answer:

Explanation:
As per mechanical energy conservation law we can say that sum of kinetic energy and potential energy will always remains constant
so here we can say

here we have

as initially it starts from rest
so now we have

now we know that its final potential energy is decreased by 10 kJ
so we have


so the final kinetic energy will be
