A force described by F(t) = ½ t2 – 5t acts on an object from t = 0 to t = 3 seconds. The object, which has a mass of 2 kg, is at rest at t = 0. Determine the velocity of the object at t = 3 seconds.
To determine the velocity of the object at t = 3 seconds, we need to calculate the integral of the force function, F(t), with respect to time from t = 0 to t = 3 seconds.
Step 1: Find the integral of F(t) = ½ t^2 - 5t with respect to time:
∫F(t) dt = ∫(½ t^2 - 5t) dt
Step 2: Evaluate the integral:
∫F(t) dt = ½ ∫t^2 dt - 5 ∫t dt
For the first integral, we use the power rule of integration:
∫t^2 dt = (1/3) t^3
For the second integral, we use the power rule again:
∫t dt = (1/2) t^2
Now we can substitute these values back into the equation:
∫F(t) dt = ½ (1/3) t^3 - 5 (1/2) t^2
Step 3: Evaluate the integral between the limits of t = 0 and t = 3:
∫F(t) dt = [½ (1/3) t^3 - 5 (1/2) t^2] evaluated from 0 to 3
Substituting the limits:
∫F(t) dt = [½ (1/3) (3^3) - 5 (1/2) (3^2)] - [½ (1/3) (0^3) - 5 (1/2) (0^2)]
Simplifying the equation:
∫F(t) dt = [½ (9) - 5 (9/2)] - [0 - 0]
Step 4: Calculate the result:
∫F(t) dt = 4.5 - 22.5 = -18
So, the integral of the force function F(t) from t = 0 to t = 3 seconds is -18.
Step 5: Apply Newton's second law to find the velocity of the object:
Newton's second law states that F = ma, where F is the net force, m is the mass of the object, and a is the acceleration. In this case, the net force is the integral of the force function, F(t), and the mass is given as 2 kg.
F = ma
-18 = 2a
Solving for acceleration, a:
a = -18/2
a = -9 m/s^2
Step 6: Calculate the velocity using the equation:
v = u + at
Since the object is at rest at t = 0, the initial velocity, u, is 0. So the equation becomes:
v = 0 + (-9)×3
v = -27 m/s
Therefore, the velocity of the object at t = 3 seconds is -27 m/s.
F(t) = 0.5t^2 - 5t = 05*3^2 - 5*3 = 10.5 N.
F = M*a.
a = F/M = 10.5/2 = 5.25 m/s^2.
V = Vo + a*t = 0 + 5.25*3 =