Suppose that an airplane accelerates at 10 m/s2 over 1000 m.

1. Suppose that an airplane accelerates at 10 m/s2 over 1000 m. Assuming it starts from rest, what is the final velocity of the airplane? (10 points)

To determine the final velocity of the airplane, we can use the following kinematic equation, which relates acceleration, distance, initial velocity, and final velocity: v2=u2+2asv^2 = u^2 + 2a s

Where:

• vv is the final velocity (what we’re solving for),
• uu is the initial velocity (which is 0 since the airplane starts from rest),
• aa is the acceleration,
• ss is the distance traveled.

Given:

• u=0 m/su = 0 \, \text{m/s},
• a=10 m/s2a = 10 \, \text{m/s}^2,
• s=1000 ms = 1000 \, \text{m}.

Substitute the given values into the equation: v2=02+2×10×1000v^2 = 0^2 + 2 \times 10 \times 1000 v2=20000v^2 = 20000

Now, take the square root of both sides to solve for vv: v=20000=141.42 m/sv = \sqrt{20000} = 141.42 \, \text{m/s}

Thus, the final velocity of the airplane is approximately 141.42 m/s.

1. If a 300 kg wagon accelerates at 1.2 m/s2 after experiencing an applied force of 625 N, what is the frictional force acting on the wagon? Assume the wagon is moving on a straight path. (10 points)

To determine the frictional force acting on the wagon, we can start by using Newton’s second law of motion: Fnet=m×aF_{\text{net}} = m \times a

Where:

• FnetF_{\text{net}} is the net force acting on the wagon,
• mm is the mass of the wagon,
• aa is the acceleration of the wagon.

Step 1: Calculate the net force

We are given:

• m=300 kgm = 300 \, \text{kg},
• a=1.2 m/s2a = 1.2 \, \text{m/s}^2.

Using Newton’s second law: Fnet=300 kg×1.2 m/s2F_{\text{net}} = 300 \, \text{kg} \times 1.2 \, \text{m/s}^2 Fnet=360 NF_{\text{net}} = 360 \, \text{N}

Step 2: Calculate the frictional force

The total applied force is 625 N, and the net force is 360 N. The net force is the result of the applied force minus the frictional force. So, we can set up the following equation: Fapplied=Ffriction+FnetF_{\text{applied}} = F_{\text{friction}} + F_{\text{net}}

Where:

• Fapplied=625 NF_{\text{applied}} = 625 \, \text{N} (the applied force),
• FfrictionF_{\text{friction}} is the frictional force.

Rearranging to solve for FfrictionF_{\text{friction}}: Ffriction=Fapplied−FnetF_{\text{friction}} = F_{\text{applied}} – F_{\text{net}} Ffriction=625 N−360 NF_{\text{friction}} = 625 \, \text{N} – 360 \, \text{N} Ffriction=265 NF_{\text{friction}} = 265 \, \text{N}

Final Answer:

The frictional force acting on the wagon is 265 N.

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