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Chapter 4: Problem 11

A boat moves through the water with two forces acting on it. One is a \(2000-\mathrm{N}\) forward push by the water on the propeller, and the other is a \(1800-\mathrm{N}\) resistive force due to the water around the bow. (a) What is the acceleration of the \(1000-\mathrm{kg}\) boat? (b) If it starts from rest, how far will the boat move in \(10.0 \mathrm{~s}\) ? (c) What will its velocity be at the end of that time?

Short Answer

Expert verified
The acceleration of the boat is \(0.2 m/s^{2}\), it will move a distance of 10m in 10 seconds, and its velocity at the end of that time will be \(2 m/s\).

Step by step solution

01

- Calculate the Net Force on the Boat

From Newton's second law, the net force on an object is equal to the mass of the object multiplied by its acceleration. But first, we need to calculate the net force. The net force acting on the boat is the difference between the forward push and the resistive force. This can be obtained by subtracting the resistive force from the forward push: \(F_{net} = F_{forward} - F_{resistive} = 2000N - 1800N = 200N \). So, the net force acting on the boat is 200N in the direction of the forward push.
02

- Find the Acceleration

Now that the net force has been found, the acceleration (a) can be calculated by rearranging Newton's second law: \(a = \frac{F_{net}}{m} = \frac{200N}{1000kg} = 0.2 m/s^{2}\). The acceleration of the boat is 0.2 m/s虏 towards the forward direction.
03

- Calculate Displacement using Equations of Motion

Now that the acceleration of the boat is known, we can find the displacement (distance travelled) using the equations of motion. Since the boat starts from rest, the initial velocity \(u = 0 m/s\) . We apply the second equation of motion: \(s = ut + \frac{1}{2}at^{2} = 0\times10s + 0.5 \times 0.2 m/s^{2} \times (10s)^{2} = 10m\). The boat will move 10m in 10s.
04

- Find the Final Velocity

The final velocity can be calculated using the first equation of motion: \(v = u + at = 0 m/s + (0.2 m/s^{2})(10s) = 2 m/s\) . Therefore, the velocity of the boat at the end of 10 s will be 2 m/s in the forward direction.

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Most popular questions from this chapter

A \(1000-\mathrm{kg}\) car is pulling a \(300-\mathrm{kg}\) trailer. Together, the car and trailer have an acceleration of \(2.15 \mathrm{~m} / \mathrm{s}^{2}\) in the positive \(x\)-direction. Neglecting frictional forces on the trailer, determine (a) the net force on the car, (b) the net force on the trailer, (c) the magnitude and direction of the force exerted by the trailer on the car, and (d) the resultant force exerted by the car on the road.

After falling from rest from a height of \(30 \mathrm{~m}\), a \(0.50-\mathrm{kg}\) ball rebounds upward, reaching a height of \(20 \mathrm{~m}\). If the contact between ball and ground lasted \(2.0 \mathrm{~ms}\), what average force was exerted on the ball?

As a fish jumps vertically out of the water, assume that only two significant forces act on it: an upward force \(F\) exerted by the tail fin and the downward force due to gravity. A record Chinook salmon has a length of \(1.50 \mathrm{~m}\) and a mass of \(61.0 \mathrm{~kg}\). If this fish is moving upward at \(3.00 \mathrm{~m} / \mathrm{s}\) as its head first breaks the surface and has an upward speed of \(6.00 \mathrm{~m} / \mathrm{s}\) after two-thirds of its length has left the surface, assume constant acceleration and determine (a) the salmon's acceleration and (b) the magnitude of the force \(F\) during this interval.

A \(5.0-g\) bullet leaves the muzzle of a rifle with a speed of \(320 \mathrm{~m} / \mathrm{s}\). What force (assumed constant) is exerted on the bullet while it is traveling down the \(0.82\)-m-long barrel of the rifle?

(a) An elevator of mass \(m\) moving upward has two forces acting on it: the upward force of tension in the cable and the downward force due to gravity. When the elevator is accelerating upward, which is greater, \(T\) or \(w ?\) (b) When the elevator is moving at a constant velocity upward, which is greater, \(T\) or \(w\) ? (c) When the elevator is moving upward, but the acceleration is downward, which is greater, \(T\) or \(w ?\) (d) Let the elevator have a mass of \(1500 \mathrm{~kg}\) and an upward acceleration of \(2.5 \mathrm{~m} / \mathrm{s}^{2}\). Find \(T\). Is your answer consistent with the answer to part (a)? (e) The elevator of part (d) now moves with a constant upward velocity of \(10 \mathrm{~m} / \mathrm{s}\). Find T. Is your answer consistent with your answer to part (b)? (f) Having initially moved upward with a constant velocity, the elevator begins to accelerate downward at \(1.50 \mathrm{~m} / \mathrm{s}^{2}\). Find \(T\). Is your answer consistent with your answer to part (c)?
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