Can someone please help with the following physics question?

Question

A block-spring system oscillates with an amplitude of 3.50 cm. The spring constant is 250 N/m and the mass of the block is 0.600 kg.

A.Determine the mechanical energy of the system in Joules
B.Determine the maximum speed of the block (m/s)
C.Determine the maximum acceleration. (m/s^2)

I would truly appreciate help with these problem.

Answer 1

mechanical energy consists of kinetic movement energy plus potential energy; and it remains constant during the motion.
in the points of maximal deviation from equilibrium point +/- L (L = 3.5 cm to the up or down the spring motion direction) the block stops, so its mechanical movement energy becomes zero, and it only has potencial energy, which is KL^2/2, where K is the spring constant, K=250 N/m.

The general equation will be mv^2/2 + Kl^2/2 = E = constant
v is velocity of the block, l is its position relatively to equilibrium point 0.
Here we have to keep in mind that v = dl/dt.

Or other equation (Using Newton's equation): mdv/dt = - Kl,
m ddl/(dt)^2 = - Kl

These equations give the same resulting motion l = L sin{sq.root(K/m)t +A}, where A is a constant sepending on what time we choose as zero.
The velocity is from here:
v = dl/dt = Lsq.root(K/m) cos{sq.root(K/m)t + A};

This gives all the question to the task:
A. The mechanic energy of the system is:
E = mv^2/2 + Kl^2/2 = mL^2K/(2m) cos^2{sq.root(K/m) + A} + KL^2/2 sin^2{sq.root(K/m) + A} = KL^2/2 ,
this is exactly the potencial energy in the maximum deviation point. (Here we used the formula sin^2 + cos^2 = 1);
E = 250 (N/m) * 0.035^2 (m^2) = 0.30625 (Nm) = 0.30625 Joules

B. The maximum speed of block is at the point, where the potencial energy equals zero (and the deviation is zero):
Vmax = +/- L sq.root(K/m)
V max = 0.035 (m) * sq.root(250 (N/m) /0.6 (kg)) = 0.035(m)*sq.root(416.67 (kg*m/(kg*m*s^2))) = 0.035 (m) * 20.41 (1/s) = 0.7144 (m/s)

C. Maximum acseleration a can be obtained from Newton equation
ma = F = Kl, and it reachs the maximum when deviation L is maximal:
a = (K/m)l (max) = KL/m = 250 (N/m)* 0.035 (m)/0.6(kg) = 14.583 m/s^2

Answer 2

A. Total Mechanical Energy is conserved throughout the motion of the block. When the block is at the aplitude A = 3.50 cm, it's speed is 0. Thus, the total energy at that point is

E = 1/2 kA^2 = 1/2 * 250 N/m * (0.035m)^2 = 0.153 J,

which is equal to the total mechanical energy of the system.

B. The maximum speed of the block is attained when there is no potential energy of the spring; namely, at x = 0. Thus, by conservation of energy, we have:

1/2 mv^2 = 0.153J

(that is, the energy of the system when x = 0 must be equal to the energy of the system at x = A, the answer to part (A) above)

=>

v = (2 * 0.153J / 0.600 kg)^1/2

=>

v = 0.714 m/s

On the other hand, the maximum acceleration occurs when the block is at the amplitude, x = A. In this case, we use Newton's second law:

F_net = ma

=>

kx = ma

=> at x = A

a = kA / m

=>

a = 250N/m * 0.035m / 0.600kg

=>

a = 14.6 m/s^2

Answer 3

A. Pe=0.5 kx^2
Pe energy of compressed spring
k - springs constant
x - max distance or amplitude A
Pe= 0.5 (250 )(3.5E-2)^2=0.153 Joules
(Stiff spring lol)

B. Ke=Pe= 0.5 mV^2 = 0.5kx^2

The max speed will occur at x=0 when potential energy becomes kinetic energy
V=x sqrt(k/m) or
V=A sqrt(k/m)
V= 3.5 E-2 sqrt(250/.6)= 0.714m/s

C. F=ma=kA (max F at x= A)

a=(k/m) (A)
a=(250/.6) (3.5E-2)=14.6 m/s^2

Answer 4

The motion of the block is given by the following equation
x = X*sinωt
where ω = sqrt(k/m) = sqrt(250 / 0.6) = 20.4 rad/s
X = 3.5 cm

v = dx/dt = X*ωcosωt
vmax = X*ω = 71.44 cm/s

a = d2x/dt2 = -X*ω^2 sinωt
a_max = Xω^2 = 1458.33 cm/s^2

Answer 5

ouch that bite. good luck . I don't quite understand the question