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A \(40 \mathrm{~kg}\) girl and an \(8.4 \mathrm{~kg}\) sled are on the frictionless ice of a frozen lake, \(15 \mathrm{~m}\) apart but connected by a rope of negligible mass. The girl exerts a horizontal \(5.2 \mathrm{~N}\) force on the rope. What are the acceleration magnitudes of (a) the sled and (b) the girl? (c) How far from the girl's initial position do they meet?

Short Answer

Expert verified
The sled accelerates at 0.619 m/s², the girl at 0.13 m/s², and they meet 2.6 m from the girl's starting point.

Step by step solution

01

Identify the Problem

We are given a girl and a sled on frictionless ice, connected by a rope. The girl exerts a force, and we need to find the acceleration of both the girl and the sled, and the meeting point from the girl's start position.
02

Define the Forces and Acceleration

Since the ice is frictionless, the force exerted by the girl on the rope causes both the girl and sled to accelerate towards each other. According to Newton's third law, the force exerted by girl and the force exerted on sled are same but in opposite directions. The formula for acceleration is given by Newton's second law: \[a = \frac{F}{m}\]Where \(F\) is force and \(m\) is mass.
03

Calculate the Sled's Acceleration

Using the formula \(a = \frac{F}{m}\), and given the sled's mass as \(8.4 \, \mathrm{kg}\) and force as \(5.2 \, \mathrm{N}\):\[a_{\text{sled}} = \frac{5.2}{8.4} = 0.619 \, \mathrm{m/s^2}\]
04

Calculate the Girl's Acceleration

Now we calculate the girl's acceleration, using her mass \(m = 40 \, \mathrm{kg}\) and the same force since actions and reactions are equal:\[a_{\text{girl}} = \frac{5.2}{40} = 0.13 \, \mathrm{m/s^2}\]
05

Calculate the Time to Meet Each Other

Both the girl and the sled are moving under constant acceleration. Since they start from rest and are accelerating towards each other, the meeting point is when their relative motion equals the initial distance \(d = 15 \, \mathrm{m}\). Using the equations of motion, the relative acceleration is:\[a_{\text{relative}} = a_{\text{sled}} + a_{\text{girl}} = 0.619 + 0.13 = 0.749 \, \mathrm{m/s^2}\]Using the equation \(d = \frac{1}{2}a t^2\):\[15 = \frac{1}{2} (0.749) t^2\]Solve for \(t\):\[t^2 = \frac{30}{0.749} \rightarrow t = \sqrt{40.053} \approx 6.33 \, \mathrm{s}\]
06

Calculate the Distance the Girl Travels

Now that we have the time \(t\), use the girl's acceleration to find the distance she travels towards the sled. Using the equation of motion \(s = \frac{1}{2} a t^2\):\[s_{\text{girl}} = \frac{1}{2} (0.13) (6.33)^2 = 2.6 \, \mathrm{m}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Frictionless Surface
A frictionless surface means that no friction forces oppose the motion of objects sliding across it. This is important for the exercise at hand because the girl and the sled move without losing momentum to friction. Friction usually converts some mechanical energy into heat, reducing the objects' acceleration. Here, the absence of friction simplifies calculations, allowing us to focus solely on the kinetic effects of the forces applied. Understanding how a frictionless surface affects motion can illuminate physics principles, like Newton's laws of motion. It helps illustrate pure motion concepts, uninhibited by real-world resistances.
  • A frictionless surface allows an object to continue moving indefinitely unless acted upon by an external force.
  • Without friction, the motion produced by a force is dictated only by Newton's second law, \( F = ma \), without reducing force effectiveness.
  • In this exercise, friction's absence means the girl and sled will accelerate according to the full force exerted on the rope.
Therefore, on a frictionless ice, the girl and sled experience pure motion, quickly highlighting concepts like momentum and acceleration due to applied force.
Acceleration Calculation
Acceleration is a critical concept in understanding motion, especially in this exercise. It tells us how swiftly an object speeds up or slows down and is calculated using Newton's second law: \( a = \frac{F}{m} \).

Finding Acceleration of the Sled

Given that the sled has a mass of \( 8.4 \, \mathrm{kg} \) and the force applied through the rope is \( 5.2 \, \mathrm{N} \), we use the formula to find the sled's acceleration:\[a_{\text{sled}} = \frac{5.2}{8.4} = 0.619 \, \mathrm{m/s^2}\]

Calculating the Girl's Acceleration

Similarly, the girl's acceleration can be determined with her mass at \( 40 \, \mathrm{kg} \):\[a_{\text{girl}} = \frac{5.2}{40} = 0.13 \, \mathrm{m/s^2}\]
  • Acceleration depends directly on force and inversely on mass. A greater force leads to greater acceleration, while a larger mass results in a smaller one for the same force applied.
  • Both the girl and sled accelerate in opposite directions due to the equality of action and reaction forces, even if their accelerations' magnitudes differ.
Understanding acceleration here showcases how mass and force interplay to affect motion differently for the girl and sled.
Relative Motion
Relative motion entails understanding how two objects move with respect to one another. In this scenario, both the girl and the sled accelerate towards each other across a frictionless surface. These movements combine to form their relative motion.The initial distance between the girl and the sled is \( 15 \, \mathrm{m} \). To find when and where they meet, we need their relative acceleration:
  • The sum of their individual accelerations determines their relative acceleration: \[a_{\text{relative}} = a_{\text{sled}} + a_{\text{girl}} = 0.619 + 0.13 = 0.749 \, \mathrm{m/s^2}\]
  • This acceleration tells us how quickly the distance between them decreases.
  • Using the equation for motion, \( d = \frac{1}{2} a t^2 \), we can calculate the time it takes for them to meet:\[15 = \frac{1}{2} (0.749) t^2\]\[t^2 = \frac{30}{0.749} \rightarrow t = \sqrt{40.053} \approx 6.33 \, \mathrm{s}\]
Once we know the time, we can find any distances covered numerically and discuss how relative motion interplays. This concept is pivotal because it shows how positions change concerning each other over time, guided by the exerted forces and surface conditions.

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Most popular questions from this chapter

A car that weighs \(1.30 \times 10^{4} \mathrm{~N}\) is initially moving at \(40 \mathrm{~km} / \mathrm{h}\) when the brakes are applied and the car is brought to a stop in \(15 \mathrm{~m}\). Assuming the force that stops the car is constant, find (a) the magnitude of that force and (b) the time required for the change in speed. If the initial speed is doubled, and the car experiences the same force during the braking, by what factors are (c) the stopping distance and (d) the stopping time multiplied? (There could be a lesson here about the danger of driving at high speeds.)

Sunjamming. A "sun yacht" is a spacecraft with a large sail that is pushed by sunlight. Although such a push is tiny in everyday circumstances, it can be large enough to send the spacecraft outward from the Sun on a cost-free but slow trip. Suppose that the spacecraft has a mass of \(900 \mathrm{~kg}\) and receives a push of \(20 \mathrm{~N}\). (a) What is the magnitude of the resulting acceleration? If the craft starts from rest, (b) how far will it travel in 1 day and (c) how fast will it then be moving?

Tarzan, who weighs \(820 \mathrm{~N}\), swings from a cliff at the end of a \(20.0 \mathrm{~m}\) vine that hangs from a high tree limb and initially makes an angle of \(22.0^{\circ}\) with the vertical. Assume that an \(x\) axis extends horizontally away from the cliff edge and a \(y\) axis extends upward. Immediately after Tarzan steps off the cliff, the tension in the vine is \(760 \mathrm{~N}\). Just then, what are (a) the force on him from the vine in unit- vector notation and the net force on him (b) in unit-vector notation and as (c) a magnitude and (d) an angle relative to the positive direction of the \(x\) axis? What are the (e) magnitude and (f) angle of Tarzan's acceleration just then?

A block with a weight of \(3.0 \mathrm{~N}\) is at rest on a horizontal surface. A \(1.0 \mathrm{~N}\) upward force is applied to the block by means of an attached vertical string. What are the (a) magnitude and (b) direction of the force of the block on the horizontal surface?

A shot putter launches a \(7.260 \mathrm{~kg}\) shot by pushing it along a straight line of length \(1.650 \mathrm{~m}\) and at an angle of \(34.10^{\circ}\) from the horizontal, accelerating the shot to the launch speed from its initial speed of \(2.500 \mathrm{~m} / \mathrm{s}\) (which is due to the athlete's preliminary motion). The shot leaves the hand at a height of \(2.110 \mathrm{~m}\) and at an angle of \(34.10^{\circ}\), and it lands at a horizontal distance of \(15.90 \mathrm{~m}\). What is the magnitude of the athlete's average force on the shot during the acceleration phase? (Hint: Treat the motion during the acceleration phase as though it were along a ramp at the given angle.)

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