A block oscillates on a spring and passes its equilibrium position with a speed of .157m/s. It's kinetic energy is zero when the block is at the distance of .1m from equilibrium. Assume no friction between the block and the table.

v=.157m/s
KE =0J when x=.1m

a)First I need to find the period of the oscillation:
1/2 mv^2=1/2 k (.1)^2
solve for m/k
m/k=.405

T=2pi*sqrt(m/k)
=2pi*sqrt(.405)
=4 s

b) what's the mass displacement from the equilibrium when its velocity v=v_mx/2?

x_max=sqrt(3/4)*(.1m)
x_max= .0866m

c) what's the spring constant if the restoring force acting on the mass when its velocity v=v_max/2 is 8.67N?

F=kx
8.67N=k(.0866m)
k=100N/m

d) what's the acceleration of the mass when v=v_max/2?:

i know that F=ma
a=F/m

Given that m/k= .405 and k=100 N/m to find m would I mulitiply .405 by 100 This would give me 40.5kg.

Finally to solve the problem a=8.67N/40.5kg= .21m/s

e)What is the KE of the mass when velocity v= v_max/2?

Isn't KE=1/2mv^2
=1.2(40.5kg)(.157m/s)^2
=.123J

f)What is the PE of the mass when velocity v= v_max/2?

PE= 1/2kx^2
=1/2(100N/m)(.0866m)^2
=.281J

Are these correct?

Let's go through each part of the problem to check if your answers are correct:

a) To find the period of the oscillation, you correctly used the formula T = 2π√(m/k). Plugging in the values m/k = 0.405 and solving gives you a period of T = 4 seconds. Your answer is correct.

b) To calculate the mass displacement from the equilibrium when the velocity is v = v_max/2, you used the equation x_max = √(3/4) * (.1m). Plugging in the values gives you x_max = 0.0866m. Your answer is correct.

c) To find the spring constant when the restoring force is 8.67N, you correctly used the equation F = kx. Plugging in the values gives you k = 100N/m. Your answer is correct.

d) To calculate the acceleration of the mass when v = v_max/2, you correctly used the equation a = F/m. Plugging in the values of F = 8.67N and m/k = 0.405, you find m = (0.405)(100) = 40.5kg. Then, plugging in the values of F = 8.67N and m = 40.5kg gives you a = 8.67N/40.5kg = 0.214m/s^2. Your answer is correct.

e) To find the kinetic energy when v = v_max/2, you used the formula KE = 1/2mv^2. However, you made a calculation error. Plugging in the values m = 40.5kg and v = 0.157m/s gives you KE = 1/2(40.5kg)(0.157m/s)^2 = 0.125J, not 0.123J. Your answer is close but slightly incorrect.

f) To calculate the potential energy when v = v_max/2, you correctly used the formula PE = 1/2kx^2. Plugging in the values k = 100N/m and x = 0.0866m gives you PE = 1/2(100N/m)(0.0866m)^2 = 0.0374J, not 0.281J. Your answer is incorrect.

In summary, your answers for parts a, b, c, and d are correct. However, your answers for parts e and f have some calculation errors.

Let's go through each step to verify the calculations:

a) To find the period of the oscillation, we use the formula T = 2π√(m/k). Given that m/k = 0.405, we can substitute this value into the formula:

T = 2π√(0.405)
T ≈ 2π(0.636)
T ≈ 4 seconds

This matches the calculation you provided.

b) To find the mass displacement from equilibrium when the velocity is v = v_max/2, we use the formula x_max = √(3/4)(x), where x is the displacement distance from equilibrium. Given that x = 0.1 m, we can substitute this value into the formula:

x_max = √(3/4)(0.1)
x_max ≈ 0.0866 m

This matches the calculation you provided.

c) To find the spring constant if the restoring force is 8.67 N, we use the formula F = kx. Given that x = 0.0866 m, we can substitute this value and the force value into the formula:

8.67 N = k(0.0866 m)
k ≈ 100 N/m

This matches the calculation you provided.

d) To find the acceleration of the mass when v = v_max/2, we use the formula a = F/m. Given that m/k = 0.405 and k = 100 N/m, we can calculate the mass:

m = (m/k) * k
m = 0.405 * 100
m = 40.5 kg

Substituting this value and the force value into the formula:

a = 8.67 N / 40.5 kg
a ≈ 0.214 m/s^2

This is slightly different from the value you calculated.

e) To find the kinetic energy of the mass when v = v_max/2, we use the formula KE = 1/2mv^2. Given that m = 40.5 kg and v = 0.157 m/s, we can substitute these values into the formula:

KE = 1/2 * 40.5 kg * (0.157 m/s)^2
KE ≈ 0.126 J

This is slightly different from the value you calculated.

f) To find the potential energy of the mass when v = v_max/2, we use the formula PE = 1/2kx^2. Given that k = 100 N/m and x = 0.0866 m, we can substitute these values into the formula:

PE = 1/2 * 100 N/m * (0.0866 m)^2
PE ≈ 0.37 J

This is different from the value you calculated.

So, some of the calculations are slightly different from the values you provided. Overall, it's important to double-check the calculations to ensure accuracy.