could somebody shows me how to get to the answer key. The answer key is 1290N.The x and y coordinates of a 4.30-kg particle moving in the xy plane under the influence of a net force F are given by
x = t4 − 5t
and
y = 5t2 + 1,
with x and y in meters and t in seconds. What is the magnitude of the force F at
t = 5.00 s?
x = t^4 - 5 t
dx/dt = 4 t^3 - 5
d^2x/dt^2 = 12 t^2 = ax
y = 5 t^2 + 1
dy/dt = 10 t
d^2y/dt^2 = 10 = ay
|a| = sqrt(100 + 144t^2)
at t = 5
|a| = sqrt (3700)
|a| = 60.8
|F| = m |a| = 4.3 * 60.8 = 261.5 N
sorry, can not get to your answer key
To find the magnitude of the force F at t = 5.00 s, we need to calculate the derivatives of the x and y equations with respect to time (t), which will give us the velocities in the x and y directions.
Let's start by finding the derivative of the x equation:
dx/dt = d(t^4 - 5t)/dt
= 4t^3 - 5
Next, let's find the derivative of the y equation:
dy/dt = d(5t^2 + 1)/dt
= 10t
Now we have the velocities in the x and y directions. To get the magnitude of the force, we need to calculate the net force using the equation:
F = ma
where F is the net force, m is the mass, and a is the acceleration.
Since the particle is moving in the xy plane, we can calculate the net force by finding the derivatives of the velocities:
dVx/dt = dvx/dt = d^2x/dt^2 = d(4t^3 - 5)/dt
= 12t^2
dVy/dt = dvy/dt = d^2y/dt^2 = d(10t)/dt
= 10
Now that we have the acceleration in the x and y directions, we can calculate the net force F:
F = m * √((dVx/dt)^2 + (dVy/dt)^2)
Given that the mass (m) is 4.30 kg, we can substitute the values into the equation:
F = 4.30 * √((12t^2)^2 + 10^2)
Now, substitute t = 5.00 s into the equation to find the magnitude of the force F:
F = 4.30 * √((12 * 5^2)^2 + 10^2)
= 4.30 * √(14400 + 100)
= 4.30 * √14500
= 4.30 * 120.415
= 517.645 N
Therefore, the magnitude of the force F at t = 5.00 s is 517.645 N.