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What force does a trampoline have to apply to a \(45.0-\mathrm{kg}\) gymnast to accelerate her straight up at \(7.50 \mathrm{~m} / \mathrm{s}^{2} ?\) Note that the answer is independent of the velocity of the gymnast-she can be moving either up or down, or be stationary.

Short Answer

Expert verified
The trampoline must apply a total force of approximately 778.95 N.

Step by step solution

01

Understand the problem

We need to calculate the force a trampoline must exert on a gymnast. We are given the mass of the gymnast, which is 45.0 kg, and the acceleration of 7.50 m/s^2. According to Newton's second law of motion, Force (F) is the product of mass (m) and acceleration (a). This can be expressed by the equation F = m * a.
02

Apply Newton's second law of motion

Using the equation F = m * a, we substitute the given mass and acceleration into the equation. So F = 45.0 kg * 7.50 m/s^2.
03

Perform the multiplication to find the force

Perform the multiplication of the mass and acceleration to find the force. F = 45.0 kg * 7.50 m/s^2 = 337.5 kg*m/s^2. Since 1 Newton (N) is equal to 1 kg*m/s^2, this means that the trampoline needs to exert a force of 337.5 N.
04

Include the force of gravity

The total force must also counteract gravity. The force of gravity on the gymnast is Fg = m * g, where g is the acceleration due to gravity (9.81 m/s^2). Fg = 45.0 kg * 9.81 m/s^2 = 441.45 N.
05

Calculate the total force the trampoline must apply

The total force the trampoline must apply is the sum of the force needed to provide the upward acceleration and the force to counteract gravity. Therefore, the total force F_total = 337.5 N + 441.45 N = 778.95 N.

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Key Concepts

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

Understanding Force Calculation
When we speak of 'force', we're referencing the push or pull exerted on an object. In the context of our exercise involving a gymnast and a trampoline, we're specifically looking at the force required to accelerate the gymnast into the air.

According to Newton's second law of motion, the force applied to an object is equal to the mass of the object multiplied by the acceleration it undergoes (\( F = m \times a \) ). To calculate this force, we simply need two pieces of information: the mass of the object and the acceleration. For our gymnast, the mass is 45.0 kg and the desired acceleration is 7.50 m/s².

The equation's beauty lies in its simplicity. Once you plug in the numbers, the calculation is straightforward. \( F = 45.0 \text{ kg} \times 7.50 \text{ m/s}^2 = 337.5 \text{ N} \). This result tells us the force that must be exerted by the trampoline, not considering other external forces just yet.
Acceleration: What Does It Mean In This Context?
Acceleration is the rate at which an object changes its velocity. It's a vector, meaning it has both a magnitude and a direction. In our textbook problem, the gymnast is being accelerated upwards, away from the ground, and towards the sky at a rate of 7.50 m/s².

It's essential to grasp that acceleration does not depend on the current velocity of the gymnast. Whether the gymnast is starting from a stationary position or already moving, the trampoline provides the same upward acceleration. This is why the gymnast's velocity doesn't affect the force calculation; all that matters are the mass and the acceleration.

This concept illustrates the universality of Newton's second law: it applies regardless of the object's state of motion, as long as you have the mass and the acceleration, you can calculate the force.
The Role of Gravitational Force
Gravity is a fundamental force that pulls objects towards each other. On Earth, it gives weight to physical objects and causes them to fall towards the ground when dropped. The gravity's acceleration is approximately \( 9.81 \text{ m/s}^2 \).

In our exercise, to calculate the force that counteracts gravity \( (F_g) \), we multiply the gymnast's mass by the acceleration due to gravity. \( F_g = 45.0 \text{ kg} \times 9.81 \text{ m/s}^2 = 441.45 \text{ N}\). This force represents the gravitational pull on the gymnast that the trampoline must overcome to achieve the desired acceleration upwards.

The total force required from the trampoline to move the gymnast includes not just the force to provide the upward acceleration but also the force to overcome gravity. Therefore, we sum these forces to find the final answer \( (F_{\text{total}} = 337.5 \text{ N} + 441.45 \text{ N} = 778.95 \text{ N}) \). Understanding gravitational force is vital since it influences so many everyday phenomena, from jumping on a trampoline to the orbits of planets.

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Most popular questions from this chapter

A powerful motorcycle can produce an acceleration of \(3.50 \mathrm{~m} / \mathrm{s}^{2}\) while traveling at \(90.0 \mathrm{~km} / \mathrm{h}\). At that speed the forces resisting motion, including friction and air resistance, total \(400 \mathrm{~N}\). (Air resistance is analogous to air friction. It always opposes the motion of an object.) What is the magnitude of the force the motorcycle exerts backward on the ground to produce its acceleration if the mass of the motorcycle with rider is \(245 \mathrm{~kg}\) ?

Integrated Concepts A basketball player jumps straight up for a ball. To do this, he lowers his body 0.300 m and then accelerates through this distance by forcefully straightening his legs. This player leaves the floor with a vertical velocity sufficient to carry him 0.900 m above the floor. (a) Calculate his velocity when he leaves the floor. (b) Calculate his acceleration while he is straightening his legs. He goes from zero to the velocity found in part (a) in a distance of 0.300 m. (c) Calculate the force he exerts on the floor to do this, given that his mass is 110 kg.

Describe a situation in which the net external force on a system is not zero, yet its speed remains constant.

Integrated Concepts A 2.50-kg fireworks shell is fired straight up from a mortar and reaches a height of 110 m. (a) Neglecting air resistance (a poor assumption, but we will make it for this example), calculate the shell’s velocity when it leaves the mortar. (b) The mortar itself is a tube 0.450 m long. Calculate the average acceleration of the shell in the tube as it goes from zero to the velocity found in (a). (c) What is the average force on the shell in the mortar? Express your answer in newtons and as a ratio to the weight of the shell.

If a constant, nonzero force is applied to an object, what can you say about the velocity and acceleration of the object?

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