Two blocks of mass m1 = 3.00 kg and m2 = 7.50 kg are connected by a mass less string that passes over a frictionless pulley. The inclines are frictionless. So this is like a triangle with a pulley at the top and both blocks resting at either side of the pulley each at a 35 degree angle.

(a) Find the magnitude of acceleration of each block.
(b) Find the tension in the string.

I found the acceleration to be 2.4 since
a=[m2sin(theta2)-m1sin(theta1)]g/(m1=m2)

For the tension I thought that I could just use m1a=T-m1gsin(theta1) but it keeps coming out wrong.

You are solving the problem correctly. Having found the acceleration of each block to be a= 2.4 mtr/sec sq, you need to feed its value in the second equation: T = m1a + m1g sin(theta1) and get the answer as T = 24.0 Nt.

34m/s^2

To find the magnitude of acceleration of each block, we can use Newton's second law.

(a) For m1, the equation is:
m1 * a = m1 * g * sin(theta1)

For m2, the equation is:
m2 * a = m2 * g * sin(theta2)

We can rearrange these equations to solve for acceleration:
a = g * sin(theta1) for m1
a = g * sin(theta2) for m2

Given that theta1 = theta2 = 35 degrees, and g = 9.8 m/s^2, we can calculate the acceleration:
a = 9.8 * sin(35 degrees)
a ≈ 5.62 m/s^2

Therefore, the magnitude of acceleration for each block is approximately 5.62 m/s^2.

(b) To find the tension in the string, we can consider the forces acting on each block.

For m1, the equation is:
T - m1 * g * sin(theta1) = m1 * a

For m2, the equation is:
m2 * g * sin(theta2) - T = m2 * a

We can substitute the known values and solve for T:
T - 3.00 kg * 9.8 m/s^2 * sin(35 degrees) = 3.00 kg * 5.62 m/s^2
T - 29.4 N = 16.9 N
T = 16.9 N + 29.4 N
T ≈ 46.3 N

Therefore, the tension in the string is approximately 46.3 N.

To find the tension in the string, we can start by analyzing the forces acting on each block separately.

For block m1, the forces involved are the weight (mg), the tension in the string (T), and the component of the weight acting along the incline (mg*sin(theta1)), where theta1 is the angle of the incline.

For block m2, the forces involved are the weight (mg), the tension in the string (T), and the component of the weight acting along the incline (mg*sin(theta2)), where theta2 is the angle of the incline.

Using Newton's second law (F = ma), we can write the following equations for each block:

For m1:
m1a = T - m1g*sin(theta1) -- Equation 1

For m2:
m2a = m2g*sin(theta2) - T -- Equation 2

Now, substitute the given values into the equations. The given masses are m1 = 3.00 kg and m2 = 7.50 kg, and let's assume the acceleration for both blocks is a (since they are connected by the same string).

For Equation 1:
3.00a = T - (3.00 kg)(9.8 m/s^2)sin(35°) -- Equation 3

For Equation 2:
7.50a = (7.50 kg)(9.8 m/s^2)sin(35°) - T -- Equation 4

To solve these equations simultaneously, we can eliminate the tension (T) term by adding Equation 3 and Equation 4:

3.00a + 7.50a = T - (3.00 kg)(9.8 m/s^2)sin(35°) + (7.50 kg)(9.8 m/s^2)sin(35°)

10.50a = (7.50 kg)(9.8 m/s^2)sin(35°) - (3.00 kg)(9.8 m/s^2)sin(35°)

10.50a = (7.50 kg - 3.00 kg)(9.8 m/s^2)sin(35°)

10.50a = (4.50 kg)(9.8 m/s^2)sin(35°)

a = (4.50 kg)(9.8 m/s^2)sin(35°) / 10.50

a ≈ 4.24 m/s^2

Now that we have determined the acceleration, we can substitute this value back into Equation 3 to find the tension (T):

3.00(4.24) = T - (3.00 kg)(9.8 m/s^2)sin(35°)

T = 3.00(4.24) + (3.00 kg)(9.8 m/s^2)sin(35°)

T ≈ 12.72 N + 29.22 N

T ≈ 41.94 N

Therefore, the tension in the string is approximately 41.94 N.