The 10-lb has an initial velocity of 10ft/s on the smooth plane. If a force F=(2.5t), where t is in seconds, acts on the block for 3s, determine the final velocity of the block and the distance the block travels during this time.
Is the F = 2.5t force in pounds? I will assume so.
The mass of the object being accelerated is W/g = 0.3106 slugs.
The final velocity after 3 seconds is the impulse divided by the mass, or
Integral (0 to 3)of F*dt/0.3106= 1.25*3^2/0.3106 = 36.2 ft/s.
Distance travelled = Integral (0 to 3) of (acceleration)*dt
acceleration = F/m = 2.5 t/0.3106 = 8.049t ft/s^2. I will leave that part up to you.
The average velocity is not half the maximum (final velocity) in this case, since the object acceleration increeases.
To solve this problem, we'll need to use Newton's second law of motion, which states that the net force acting on an object is equal to the object's mass times its acceleration (F = ma).
Given:
- Mass of the block (m) = 10 lb
- Initial velocity (v0) = 10 ft/s
- Force (F) = 2.5t
- Time (t) = 3 seconds
To find the final velocity (v) of the block after 3 seconds, we'll first calculate the acceleration (a) by dividing the force by the mass:
a = F / m
a = (2.5t) / 10
a = 0.25t
Next, we'll integrate the acceleration function with respect to time to find the velocity function:
v = ∫(0.25t) dt
v = 0.25 * ∫t dt
v = 0.25 * (t^2 / 2) = 0.125t^2
Now, we'll substitute the value of time (t = 3) into the velocity function to find the final velocity:
v = 0.125 * (3^2)
v = 0.125 * 9
v = 1.125 ft/s
Therefore, the final velocity of the block after 3 seconds is 1.125 ft/s.
To determine the distance traveled by the block during this time, we can use the kinematic equation:
d = v0 * t + (1/2) * a * t^2
d = 10 * 3 + (1/2) * 0.25 * 3^2
d = 30 + (1/2) * 0.25 * 9
d = 30 + 0.375
d = 30.375 ft
Hence, the block travels a distance of 30.375 ft during the 3-second interval.
To determine the final velocity and distance traveled by the block, we can use the equations of motion.
Step 1: Calculate the acceleration.
The force acting on the block is given by F = m * a, where F is the force, m is the mass, and a is the acceleration. In this case, the mass of the block is 10 lbs.
Since the force is given by F = 2.5t, we can equate it to m * a: 2.5t = 10 * a.
Step 2: Solve for acceleration (a).
Rearranging the equation, we have a = 2.5t / 10.
Step 3: Integrate to find the velocity.
The velocity is the integral of acceleration with respect to time. Integrating the equation a = 2.5t / 10 with respect to t gives us:
v = ∫(2.5t / 10) dt.
Integrating, we get v = (2.5/10) * (t^2/2) + C,
where C is the constant of integration.
Step 4: Plug in the initial conditions.
Since the initial velocity is given as 10 ft/s, we can substitute v = 10 when t = 0 into the above equation to solve for C.
10 = (2.5/10) * (0^2/2) + C,
10 = 0 + C,
C = 10.
Step 5: Substitute C into the equation of the velocity.
Thus, the equation for the velocity becomes: v = (2.5/10) * (t^2/2) + 10.
Step 6: Calculate the final velocity.
To determine the final velocity, substitute t = 3 into the equation:
v = (2.5/10) * (3^2/2) + 10.
Calculating, we get v = 3.375 ft/s (rounded to three decimal places).
Step 7: Calculate the distance traveled.
To find the distance traveled by the block during this time, we need to integrate the velocity with respect to time.
d = ∫v dt = ∫[(2.5/10) * (t^2/2) + 10] dt.
Integrating, we get d = (2.5/10) * (t^3/6) + 10t + C2,
where C2 is the constant of integration.
Step 8: Plug in the initial conditions.
Since the initial velocity is given as 10 ft/s, we can substitute d = 0 when t = 0 into the above equation to solve for C2.
0 = (2.5/10) * (0^3/6) + 10 * 0 + C2,
0 = 0 + 0 + C2,
C2 = 0.
Step 9: Substitute C2 into the equation of the distance.
Thus, the equation for the distance becomes: d = (2.5/10) * (t^3/6) + 10t.
Step 10: Calculate the distance traveled.
To find the distance traveled during the 3 seconds, substitute t = 3 into the equation:
d = (2.5/10) * (3^3/6) + 10 * 3.
Calculating, we get d = 18.75 ft (rounded to two decimal places).
Therefore, the final velocity of the block is 3.375 ft/s, and the distance traveled during the 3-second duration is 18.75 ft.