A 2.9 kg body is at rest on a frictionless horizontal air track when a horizontal force F acting in the positive direction of an x axis along the track is applied to the body. A stroboscopic graph of the position of the body as it slides to the right is shown in the figure. The force F is applied to the body at t = 0, and the graph records the position of the body at 0.50 s intervals. How much work is done on the body by the applied force F between t = 0 and t = 1.0 s? (HINT: Find the average acceleration over t = 0 to t = 1.0 s.)
The figure shows the following data:
t=0 x(position)=0
t=.5 x=.04
t=1.0 x=.2
t=1.5 x=.44
t=2.0 x=.8
To find the work done on the body by the applied force between t=0 and t=1.0 s, we will first need to calculate the average acceleration of the body over that time interval.
The average acceleration can be calculated using the equation:
average acceleration = (change in velocity) / (time interval)
In this case, the time interval is 1.0 s, so we need to calculate the change in velocity over that time interval.
From the provided data, we can see that the position of the body at t=0 is 0, and at t=1.0 s is 0.2. To find the change in velocity, we need to find the change in position.
change in position = final position - initial position
change in position = 0.2 - 0
change in position = 0.2 m
Since the velocity is the rate of change of position with respect to time, we can write:
change in velocity = (change in position) / (time interval)
change in velocity = (0.2 m) / (1.0 s)
change in velocity = 0.2 m/s
Now that we have the change in velocity, we can calculate the average acceleration.
average acceleration = (change in velocity) / (time interval)
average acceleration = (0.2 m/s) / (1.0 s)
average acceleration = 0.2 m/s^2
Now, we can calculate the work done on the body by the applied force using the equation:
work done = force * displacement * cos(angle)
Since the force is acting in the positive direction of the x-axis and the displacement is positive (from the initial position at t=0 to the final position at t=1.0 s), the angle between the force and displacement is 0 degrees. Therefore, the cosine of the angle will be 1.
work done = force * displacement
We are given the mass of the body (2.9 kg), but we still need to find the force.
Using Newton's second law (F = ma), we can find the force by multiplying the mass of the body with the average acceleration:
force = mass * average acceleration
force = 2.9 kg * 0.2 m/s^2
force = 0.58 N
Now we can calculate the work done:
work done = force * displacement
work done = 0.58 N * 0.2 m
work done = 0.116 J
Therefore, the work done on the body by the applied force F between t = 0 and t = 1.0 s is 0.116 Joules.
To find the work done on the body by the applied force F between t = 0 and t = 1.0 s, we need to calculate the average acceleration over that time interval.
The average acceleration can be determined by using the formula:
average acceleration = (change in velocity) / (time interval)
Since the body is at rest initially, the initial velocity is 0. Therefore, the change in velocity will be the final velocity.
To calculate the final velocity, we can use the formula for acceleration:
acceleration = (change in velocity) / (time interval)
Rearranging the formula, we get:
change in velocity = acceleration * time interval
Then, we can substitute the values given for the position and time in the stroboscopic graph:
change in velocity = (0.2 - 0) / 1.0 = 0.2 m/s
Now that we have the change in velocity, we can calculate the acceleration:
acceleration = change in velocity / time interval = 0.2 m/s / 1.0 s = 0.2 m/s²
Next, we can calculate the work done on the body using the formula:
work = force * displacement * cos(theta)
Since the force is applied in the positive x direction and the displacement is also in the positive x direction, the angle between them (theta) is 0 degrees (cos(0) = 1).
The force can be calculated using Newton's second law:
force = mass * acceleration = 2.9 kg * 0.2 m/s² = 0.58 N
Finally, we can calculate the work done:
work = 0.58 N * 0.2 m * cos(0) = 0.116 J
Therefore, the work done on the body by the applied force F between t = 0 and t = 1.0 s is 0.116 Joules.