A 4.90 kg particle moves along the x axis. Its position varies with time according to x = t + 3.4t3, where x is in meters and t is in seconds.
(a) Find the kinetic energy at any time t. (Accurately round any coefficient to exactly two decimal places. Use t as necessary.)
K = . J
(b) Find the acceleration of the particle and the force acting on it at time t. (Accurately round any coefficient to exactly two decimal places. Use t as necessary.)
a
= m/s2
F
= N
(c) Find the power being delivered to the particle at time t. (Accurately round any coefficient to exactly two decimal places. Use t as necessary.)
P = W
(d) Find the work done on the particle in the interval t = 0 to t = 2.00 s.
J
To find the answers to these questions, we need to differentiate the given position function to find velocity and acceleration.
Given position function: x = t + 3.4t^3
(a) Kinetic energy (K) can be found using the equation: K = (1/2)mv^2, where m is the mass and v is the velocity.
We first need to find the velocity by differentiating the position function with respect to time (t).
Taking the derivative of x with respect to t, we get:
v = dx/dt = 1 + 10.2t^2
Now, we can substitute the mass (m = 4.90 kg) and the velocity (v = 1 + 10.2t^2) into the kinetic energy equation:
K = (1/2) * 4.90 * (1 + 10.2t^2)^2
Simplifying the equation and accurately rounding any coefficients to two decimal places, we get:
K = 2.45(1 + 20.4t^2 + 104.04t^4) J
(b) Acceleration (a) can be found by taking the derivative of velocity with respect to time (t).
Taking the derivative of v with respect to t, we get:
a = dv/dt = 20.4t
To find the force (F) acting on the particle, we use Newton's second law: F = ma.
Substituting the mass (m = 4.90 kg) and the acceleration (a = 20.4t) into the equation, we get:
F = 4.90 * 20.4t
Simplifying the equation and accurately rounding any coefficients to two decimal places, we get:
F = 98.86t N
(c) Power (P) can be found using the equation: P = F * v, where F is the force and v is the velocity.
Substituting the force (F = 98.86t) and the velocity (v = 1 + 10.2t^2) into the equation, we get:
P = (98.86t)(1 + 10.2t^2)
Simplifying the equation and accurately rounding any coefficients to two decimal places, we get:
P = 98.86t + 1012.57t^3 W
(d) Work done (W) can be found using the equation: W = ∫F dx, where F is the force and dx is the change in position.
Substituting the force (F = 98.86t) into the equation, we get:
W = ∫(98.86t) dx
Integrating the equation with respect to x from x = 0 to x = t, we get:
W = 98.86 * ∫t dx = 98.86 * (t^2/2)
Substituting the time interval t = 2.00 s, we get:
W = 98.86 * (2.00^2/2) J
Calculating the result, we get:
W = 197.72 J