Questions: A 2 kg skateboard is rolling across a smooth, flat floor when a girl kicks it, causing it to speed up to 4.5 m / s in 0.5 seconds without changing direction. If the average force exerted by the girl on the skateboard in its direction of motion was 6 N, with what initial velocity was it moving?

A 2 kg skateboard is rolling across a smooth, flat floor when a girl kicks it, causing it to speed up to 4.5 m / s in 0.5 seconds without changing direction. If the average force exerted by the girl on the skateboard in its direction of motion was 6 N, with what initial velocity was it moving?
Transcript text: 8. A 2 kg skateboard is rolling across a smooth, flat floor when a girl kicks it, causing it to speed up to $4.5 \mathrm{~m} / \mathrm{s}$ in 0.5 seconds without changing direction. If the average force exerted by the girl on the skateboard in its direction of motion was 6 N , with what initial velocity was it moving?
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Solution

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Solution Steps

Step 1: Understand the Problem

We need to find the initial velocity of the skateboard. We know the final velocity, the time over which the force was applied, and the average force exerted. We can use the principles of physics, specifically Newton's second law and the equations of motion, to solve this problem.

Step 2: Use Newton's Second Law

Newton's second law states that the force exerted on an object is equal to the mass of the object multiplied by its acceleration:

\[ F = m \cdot a \]

Given:

  • \( F = 6 \, \mathrm{N} \)
  • \( m = 2 \, \mathrm{kg} \)

We can solve for acceleration \( a \):

\[ a = \frac{F}{m} = \frac{6 \, \mathrm{N}}{2 \, \mathrm{kg}} = 3 \, \mathrm{m/s^2} \]

Step 3: Use the Equation of Motion

The equation of motion that relates initial velocity \( v_i \), final velocity \( v_f \), acceleration \( a \), and time \( t \) is:

\[ v_f = v_i + a \cdot t \]

Given:

  • \( v_f = 4.5 \, \mathrm{m/s} \)
  • \( a = 3 \, \mathrm{m/s^2} \)
  • \( t = 0.5 \, \mathrm{s} \)

We can solve for the initial velocity \( v_i \):

\[ v_i = v_f - a \cdot t = 4.5 \, \mathrm{m/s} - 3 \, \mathrm{m/s^2} \cdot 0.5 \, \mathrm{s} \]

\[ v_i = 4.5 \, \mathrm{m/s} - 1.5 \, \mathrm{m/s} = 3.0 \, \mathrm{m/s} \]

Final Answer

The initial velocity of the skateboard was \(\boxed{3.0 \, \mathrm{m/s}}\).

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