Question

A student witnesses an 8.0-kilogram shopping cart undergoing uniform acceleration across a parking lot. It achieves a velocity of 4.0 meters per second in 0.5 seconds. What is the force's magnitude responsible for the acceleration of the shopping cart?

Answer 1

To find the force responsible for the acceleration of the shopping cart, we can use Newton's second law of motion, which states that \( F = ma \), where \( F \) is the force, \( m \) is the mass, and \( a \) is the acceleration.

1. **Determine the acceleration** \( a \):
We can use the formula for acceleration, which is given by:
\[
a = \frac{\Delta v}{\Delta t}
\]
where \( \Delta v \) is the change in velocity and \( \Delta t \) is the time taken. The change in velocity is:
\[
\Delta v = v_f - v_i = 4.0 \, \text{m/s} - 0 \, \text{m/s} = 4.0 \, \text{m/s}
\]
The time taken is \( \Delta t = 0.5 \, \text{s} \).

Now we can calculate the acceleration:
\[
a = \frac{4.0 \, \text{m/s}}{0.5 \, \text{s}} = 8.0 \, \text{m/s}^2
\]

2. **Calculate the force** \( F \):
We now use the mass of the shopping cart, which is \( m = 8.0 \, \text{kg} \), to find the force:
\[
F = m \cdot a
\]
Substituting the values we have:
\[
F = 8.0 \, \text{kg} \cdot 8.0 \, \text{m/s}^2 = 64.0 \, \text{N}
\]

Therefore, the magnitude of the force responsible for the acceleration of the shopping cart is \( \boxed{64.0} \, \text{N} \).