An electric motor is used to pull a 125 kg box across a floor using a long cable. the tension in the cable is 350N and the box accelerates at 1.2m/s^2[forward] for 5.0s. The cable breaks and the box slows down and stops.
How far does the box travel from the moment the cable breaks until it stops?
To find the distance the box travels from the moment the cable breaks until it stops, we can use the equation of motion:
\[ d = v_i t + \frac{1}{2} a t^2 \]
where:
- d is the distance traveled
- \( v_i \) is the initial velocity
- a is the acceleration
- t is the time
Given:
- \( v_i = 1.2 \, \text{m/s} \) (initial velocity after the cable breaks)
- a = -1.2 m/s^2 (acceleration is negative as the box is slowing down)
- t = 5.0 s (time until the box stops)
Plugging in these values into the equation, we get:
\[ d = (1.2 \, \text{m/s}) \cdot (5.0 \, \text{s}) + \frac{1}{2} (-1.2 \, \text{m/s}^2) \cdot (5.0 \, \text{s})^2 \]
Simplifying the equation, we have:
\[ d = 6 \, \text{m} - \frac{1}{2} (1.2 \, \text{m/s}^2) \cdot (25 \, \text{s}^2) \]
Calculating further:
\[ d = 6 \, \text{m} - \frac{1}{2} (1.2 \, \text{m/s}^2) \cdot (625 \, \text{m}^2/\text{s}^2) \]
\[ d = 6 \, \text{m} - 0.6 \, \text{m} \cdot \text{s}^2 \]
Therefore, the box travels a distance of:
\[ d = 5.4 \, \text{m} \]
To find the distance the box travels from the moment the cable breaks until it stops, we need to calculate the distance traveled during the acceleration phase and the distance traveled during the deceleration phase.
During the acceleration phase, the box experiences a net force in the forward direction, given by Newton's second law of motion:
net force = mass × acceleration
In this case, the net force is the tension in the cable, which is 350 N, the mass of the box is 125 kg, and the acceleration is 1.2 m/s^2. Therefore, we can calculate the net force as follows:
350 N = 125 kg × 1.2 m/s^2
Next, we can use the formula for calculating distance during uniform acceleration:
distance = (initial velocity × time) + (0.5 × acceleration × time^2)
During the acceleration phase, the initial velocity is zero since the box starts from rest. The time is 5.0 seconds, and the acceleration is 1.2 m/s^2. Plugging in these values, we can calculate the distance traveled during the acceleration phase:
distance = (0 × 5.0) + (0.5 × 1.2 × (5.0^2))
distance = 0 + (0.5 × 1.2 × 25.0)
distance = 0 + (0.5 × 30.0)
distance = 0 + 15.0
distance = 15.0 meters
Therefore, the box travels 15.0 meters during the acceleration phase.
Now, during the deceleration phase, the box is slowing down due to the absence of tension in the cable. The deceleration is the opposite direction of acceleration, which means it has the same magnitude but is in the opposite direction. So the deceleration is -1.2 m/s^2.
Using the same formula for distance during uniform acceleration, with a negative acceleration value:
distance = (initial velocity × time) + (0.5 × acceleration × time^2)
During the deceleration phase, the initial velocity is the velocity at the moment the cable breaks. We don't have this value directly, but we can calculate it by using the equation:
final velocity = initial velocity + (acceleration × time)
At the moment the cable breaks, the box has been accelerating for 5.0 seconds, so the final velocity is:
final velocity = 0 m/s + (1.2 m/s^2 × 5.0 s)
final velocity = 0 m/s + 6.0 m/s
final velocity = 6.0 m/s
Now, we can calculate the distance traveled during the deceleration phase:
distance = (initial velocity × time) + (0.5 × acceleration × time^2)
Plugging in the values, we have:
distance = (6.0 m/s × t) + (0.5 × -1.2 m/s^2 × t^2)
Since the box slows down and comes to a stop, the final velocity is 0 m/s. We can use this information to find the total time it takes for the box to decelerate to 0 m/s:
0 m/s = 6.0 m/s - (1.2 m/s^2 × t)
1.2 m/s^2 × t = 6.0 m/s
t = 6.0 m/s / 1.2 m/s^2
t = 5.0 s
Now, substituting the time back into the distance formula, we can find the distance traveled during the deceleration phase:
distance = (6.0 m/s × 5.0 s) + (0.5 × -1.2 m/s^2 × (5.0^2))
distance = 30.0 m - 15.0 m
distance = 15.0 m
Therefore, the box travels an additional 15.0 meters during the deceleration phase.
To find the total distance traveled from the moment the cable breaks until the box stops, we add the distances traveled during the acceleration and deceleration phases:
total distance = distance during acceleration + distance during deceleration
total distance = 15.0 m + 15.0 m
total distance = 30.0 meters
Therefore, the box travels a total distance of 30.0 meters from the moment the cable breaks until it stops.