I cant figure this problem out and it's asking for the value and units for the force. Please help me!

A 5.25 kg crate is suspended from the end of a short vertical rope of negligible mass. An upward force F(t) is applied to the end of the rope, and the height of the crate above its initial position is given by y(t)=(2.80m/s)t +(0.61(m/s^3)(t^3).

-What is the magnitude of the force F when 3.60 s?

To find the magnitude of the force F at a given time, we need to determine the value of the force F at that time.

The height of the crate above its initial position, y(t), is given by the equation y(t) = (2.80 m/s)t + (0.61 (m/s^3))(t^3).

To find the force at a specific time, we need to take the derivative of the height equation with respect to time (t):

dy/dt = (d/dt)(2.80 t) + (d/dt)(0.61 t^3)
dy/dt = 2.80 + 3(0.61)(t^2)
dy/dt = 2.80 + 1.83t^2

The derivative of the height equation represents the velocity of the crate at a given time. Since the force F(t) is equal to the mass of the crate (5.25 kg) multiplied by the acceleration, we can find the force at a specific time by multiplying the mass by the acceleration.

The acceleration can be found by taking the second derivative of the height equation with respect to time:

d^2y/dt^2 = (d/dt)(2.80 + 1.83t^2)
d^2y/dt^2 = 0 + 2(1.83)(t)
d^2y/dt^2 = 3.66t

Now we can find the acceleration of the crate at a specific time by plugging in the value of t = 3.60 s:

a(3.60) = 3.66(3.60)
a(3.60) = 13.176 m/s^2

Finally, we can find the magnitude of the force by multiplying the mass of the crate by the acceleration:

F = m * a
F = 5.25 kg * 13.176 m/s^2

Calculating the value of the force:

F ≈ 68.97 N

Therefore, the magnitude of the force F at t = 3.60 s is approximately 68.97 N.

To find the magnitude of the force F at the given time, we need to remember the relationship between force and motion.

In this problem, the force F(t) is responsible for the crate's upward motion, resulting in a change in height over time. The height of the crate is given by the equation y(t) = (2.80 m/s)t + (0.61 (m/s^3))(t^3).

To determine the force, we need to use Newton's second law, which states that force (F) is equal to mass (m) multiplied by acceleration (a). In this case, acceleration is the second derivative of the height function y(t) with respect to time (t). Therefore, we need to take the equation for y(t) and differentiate it twice to find the acceleration.

First, differentiate the equation once to find the velocity (v):
v(t) = dy(t)/dt = 2.80 m/s + 3(0.61 (m/s^3))(t^2) = 2.80 m/s + 1.83 (m/s^3)(t^2)

Then, differentiate the velocity equation once more to find the acceleration (a):
a(t) = dv(t)/dt = d^2y(t)/dt^2 = 0 + 2(1.83 (m/s^3))(t) = 3.66 (m/s^3)t

Now that we have the acceleration, we can determine the force F using Newton's second law:
F(t) = m * a(t) = 5.25 kg * (3.66 (m/s^3)t)

Now, to find the magnitude of the force F at t = 3.60 s, we plug in the value of t into the equation for F:
F(3.60 s) = 5.25 kg * (3.66 (m/s^3))(3.60 s)

Solving this equation will give you the value and units for the force at t = 3.60 s.

F = m a

what is a?
d^2y/dt^2 = a
dy/dt = 2.80 +(3*.61)t^2
d^2y/dt^2 = a = (6*.61) t
F = m a = (5.25*6*.61)t
so when t = 3.6
F = 5.25 * 6 * .61 * 3.6 Newtons