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As shown below, if \(F=60.0 \mathrm{N}\) and \(M=4.00 \mathrm{kg}\) what is the magnitude of the acceleration of the suspended object? All surfaces are frictionless.

Short Answer

Expert verified
The magnitude of the acceleration of the suspended object is \(15 \, m/s^2\).

Step by step solution

01

Identify and Organize

First, identify the given variables. In this problem, the force \(F\) acting on the object is 60N and the mass \(M\) of the object is 4kg. We're tasked with finding the acceleration, which can be calculated from the given variables using Newton's second law.
02

Apply Newton's Second Law

Next, apply Newton's second law of motion, which states that the acceleration of an object is equal to the net force acting on it divided by its mass. Algebraically, this law can be expressed as \(a = F/m\).
03

Calculate the acceleration

Replace the force and mass values into the equation: \(a = 60N / 4kg\). The unit of force (the newton) is a compound unit which can be mathematically broken down into a kilogram meter per second squared (\(kg \cdot m/s^2\)). Therefore, dividing newtons by kilograms, we are left with meters per second squared, which is a unit of acceleration.
04

Final Calculation

Performing the division gives a solution of \(a = 15 \, m/s^2\). Therefore, the acceleration of the object is \(15 \, m/s^2\).

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Key Concepts

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

Force
In physics, "force" is a fundamental concept that describes an interaction capable of changing an object's state of motion or rest. Technically, force is a vector quantity, which means it has both magnitude and direction. Imagine you are pushing a shopping cart. The force you apply makes the cart move faster, change direction, or come to a stop.

**Newton's Second Law**
Newton's Second Law of Motion connects force with mass and acceleration, written as: \[F = ma\] Where:
  • \( F \) stands for force, measured in newtons (N).
  • \( m \) represents mass, measured in kilograms (kg).
  • \( a \) denotes acceleration, measured in meters per second squared ( m/s^2 ).
For example, when the force acting on an object is 60 N and the object's mass is 4 kg, you can determine how quickly it speeds up by calculating the acceleration using the formula above.
Mass
Mass is a measure of the amount of matter in an object and is typically measured in kilograms (kg). It tells us how much an object resists changing its motion. In everyday terms, mass is often confused with weight, but remember that weight is a force, whereas mass is the amount of matter.

**Mass and Inertia**
In physical terms, mass is closely related to inertia, which refers to an object's resistance to changes in its state of motion. The greater the mass, the greater the inertia and the more force is needed to change its motion.

**Mass in Newton's Second Law**
In Newton's Second Law, mass plays a crucial role. It helps determine the acceleration of an object when a specific force is applied. Larger masses require more force to achieve the same acceleration as smaller masses. In the given problem, we have \( m = 4 \, \text{kg} \), which means the object's motion will change based on this mass when a force is exerted on it.
Acceleration
Acceleration refers to the rate of change of an object's velocity. It measures how quickly an object speeds up, slows down, or changes direction. The standard unit for acceleration is meters per second squared ( m/s^2 ).

**Understanding Acceleration**
Acceleration occurs whenever there is a change in velocity. Think about a car increasing its speed when you press the gas pedal—this is acceleration.

**Calculating Acceleration**
Using Newton's Second Law, you can find the acceleration by rearranging the formula:\[a = \frac{F}{m}\]For example, if a force of 60 N acts on a 4 kg object, you can calculate its acceleration as follows:\[a = \frac{60 \, \text{N}}{4 \, \text{kg}} = 15 \, m/s^2\]This means the object will accelerate at a rate of 15 meters per second squared. Understanding acceleration helps in visualizing how quickly an object's movement alters under applied forces.

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Most popular questions from this chapter

A roller coaster car starts from rest at the top of a track \(30.0 \mathrm{m}\) long and inclined at \(20.0^{\circ}\) to the horizontal. Assume that friction can be ignored. (a) What is the acceleration of the car? (b) How much time elapses before it reaches the bottom of the track?

. A car of mass \(1000.0 \mathrm{kg}\) is traveling along a level road at \(100.0 \mathrm{km} / \mathrm{h}\) its brakes are applied. Calculate the stopping distance if the coefficient of kinetic friction of the tires is 0.500. Neglect air resistance. (Hint: since the distance traveled is of interest rather than the time, \(x\) is the desired independent variable and not \(t .\) Use the Chain Rule to change the variable: \(\frac{d v}{d t}=\frac{d v}{d x} \frac{d x}{d t}=v \frac{d v}{d x}\)

A skydiver is at an altitude of 1520 m. After 10.0 seconds of free fall, he opens his parachute and finds that the air resistance, \(F_{D},\) is given by the formula \(F_{\mathrm{D}}=-b v, \quad\) where \(b\) is a constant and \(v\) is the velocity. If \(b=0.750,\) and the mass of the skydiver is \(82.0 \mathrm{kg}\) first set up differential equations for the velocity and the position, and then find: (a) the speed of the skydiver when the parachute opens, (b) the distance fallen before the parachute opens, (c) the terminal velocity after the parachute opens (find the limiting velocity), and (d) the time the skydiver is in the air after the parachute opens.

When a body of mass 0.25 kg is attached to a vertical massless spring, it is extended \(5.0 \mathrm{cm}\) from its unstretched length of \(4.0 \mathrm{cm} .\) The body and spring are placed on a horizontal frictionless surface and rotated about the held end of the spring at 2.0 rev/s. How far is the spring stretched?

Modern roller coasters have vertical loops like the one shown here. The radius of curvature is smaller at the top than on the sides so that the downward centripetal acceleration at the top will be greater than the acceleration due to gravity, keeping the passengers pressed firmly into their seats. (a) What is the speed of the roller coaster at the top of the loop if the radius of curvature there is \(15.0 \mathrm{m}\) and the downward acceleration of the car is \(1.50 \mathrm{g}\) ? (b) How high above the top of the loop must the roller coaster start from rest, assuming negligible friction? (c) If it actually starts \(5.00 \mathrm{m}\) higher than your answer to (b), how much energy did it lose to friction? Its mass is \(1.50 \times 10^{3} \mathrm{kg}\)

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