To solve the problem of how far a 35.0 kg object would travel when a 355 N unbalanced force is applied for 6.00 seconds, we need to proceed using the following steps:
### Step 1: Identify Given Data
- Force (F) = 355 Newtons (N)
- Mass (m) = 35.0 kilograms (kg)
- Time (t) = 6.00 seconds (s)
### Step 2: Identify What We Need to Find
We need to find the distance (d) the object travels in 6.00 seconds.
### Step 3: Calculate the Acceleration
Using Newton's second law of motion:
[tex]\[ F = m \cdot a \][/tex]
where:
- [tex]\( F \)[/tex] is the force applied,
- [tex]\( m \)[/tex] is the mass of the object,
- [tex]\( a \)[/tex] is the acceleration.
Rearranging the formula to solve for acceleration ([tex]\( a \)[/tex]):
[tex]\[ a = \frac{F}{m} \][/tex]
Substitute the given values:
[tex]\[ a = \frac{355 \text{ N}}{35.0 \text{ kg}} \][/tex]
[tex]\[ a = 10.142857142857142 \text{ m/s}^2 \][/tex]
### Step 4: Calculate the Distance Traveled
Since the object starts from rest (initial velocity [tex]\( u = 0 \)[/tex]), we can use the kinematic equation:
[tex]\[ d = ut + \frac{1}{2} a t^2 \][/tex]
Given that [tex]\( u = 0 \)[/tex], the equation simplifies to:
[tex]\[ d = \frac{1}{2} a t^2 \][/tex]
Substitute the calculated acceleration and the given time:
[tex]\[ d = \frac{1}{2} \times 10.142857142857142 \text{ m/s}^2 \times (6.00 \text{ s})^2 \][/tex]
[tex]\[ d = 0.5 \times 10.142857142857142 \times 36 \][/tex]
[tex]\[ d = 182.57142857142856 \text{ meters} \][/tex]
### Step 5: Conclusion
After 6.00 seconds, the object would travel a distance of approximately 182.57 meters.
