a horizontal spring is stretched from its equilibrium position when a force of 0.75kg acts on it. then a force of 1.5 kg is attached to the end of the end of the spring and pulled 4.5 cm along a frictionless table from the equilibrium position. the body is then released and performs a simple harmonic motion calculate (1) the force constant of the spring (2)force exerted by the spring on the 1.5kg body before it was released?

working
f=ke
0.75+1.5=2.25kg
2.25kg*10 =22.5n
k =f/e= 22.5n/0.045m= 500nm
but I checked the answer from the answer page my answers were wrong please help oo.

To calculate the force constant of the spring (k), you need to use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position.

Given:
Force exerted on the spring = 0.75 kg
Displacement of the spring = 4.5 cm = 0.045 m

To find k, the force constant, use Hooke's Law: F = k * x, where F is the force exerted on the spring (0.75 kg * 10 m/s^2 = 7.5 N) and x is the displacement of the spring (0.045 m). So we have:

k * 0.045 m = 7.5 N

Therefore, the force constant (k) is:

k = 7.5 N / 0.045 m = 166.67 N/m

Now, to calculate the force exerted by the spring on the 1.5 kg body before it was released, you can use the equation F = k * x, where F is the force exerted by the spring, k is the force constant (166.67 N/m), and x is the displacement of the spring (0.045 m).

So, the force exerted by the spring on the 1.5 kg body before it was released is:

F = 166.67 N/m * 0.045 m = 7.5 N

Note: In your working, you mistakenly added 0.75 kg and 1.5 kg to get 2.25 kg, which is incorrect. You should have added 0.75 kg and 1.5 kg to obtain the total mass being acted upon by the spring, which is 2.25 kg. Additionally, you mistakenly multiplied the total mass (2.25 kg) by 10 m/s^2, which is incorrect. The gravitational force is not relevant here since the problem specifies that the spring is stretched horizontally on a frictionless table.