Below is a conservation of energy problem. The solution to this problem is provided. Assess whether the solution provided is correct or incorrect AND EXPLAIN WHY. 4.6 kg block is at rest against a horizontal spring that is compressed by 0.40 m. The spring has a spring constant of k₁ = 2750 N/m. After leaving the spring, it travels up a 28° incline to a height of 2.8 m. At the top of the hill is a second spring with a spring constant of k₂ = 350 N/m. The horizontal portions are frictionless, but the hill has a coefficient of kinetic friction equal to uk= 0.16. The final velocity of the block is 9.78 m/s. How much is the second spring compressed by when the block comes to a stop against it? Simplifies to X₁ = 0.40 m h₁ = 0 m Vi = 0 m/s X = K₁ = 2750 N/m K₂= Wnc = 0 J 2 2 2 Wnc + mghi +½ kx₁² +½ mv² = mgh₁+½ kxf² +½ mvf2 Xf= = 350 N/m m = 4.6 kg 2 1/2 k₁x₁² = mghf + ¼ K₂Xf2 2750 N/m 350 N/m 2 ½ k₁ס² — mgħ –½ k₂xf² X (0.4 m)2 k 2 -X k₂ i f= 2gh 0.141 m X₁ = ? m hf = 2.8 m Vi = 0 m/s f 2(9.8 m/s2)(2.8 m) 2750 N/m
Kinematics
A machine is a device that accepts energy in some available form and utilizes it to do a type of work. Energy, work, or power has to be transferred from one mechanical part to another to run a machine. While the transfer of energy between two machine parts, those two parts experience a relative motion with each other. Studying such relative motions is termed kinematics.
Kinetic Energy and Work-Energy Theorem
In physics, work is the product of the net force in direction of the displacement and the magnitude of this displacement or it can also be defined as the energy transfer of an object when it is moved for a distance due to the forces acting on it in the direction of displacement and perpendicular to the displacement which is called the normal force. Energy is the capacity of any object doing work. The SI unit of work is joule and energy is Joule. This principle follows the second law of Newton's law of motion where the net force causes the acceleration of an object. The force of gravity which is downward force and the normal force acting on an object which is perpendicular to the object are equal in magnitude but opposite to the direction, so while determining the net force, these two components cancel out. The net force is the horizontal component of the force and in our explanation, we consider everything as frictionless surface since friction should also be calculated while called the work-energy component of the object. The two most basics of energy classification are potential energy and kinetic energy. There are various kinds of kinetic energy like chemical, mechanical, thermal, nuclear, electrical, radiant energy, and so on. The work is done when there is a change in energy and it mainly depends on the application of force and movement of the object. Let us say how much work is needed to lift a 5kg ball 5m high. Work is mathematically represented as Force ×Displacement. So it will be 5kg times the gravitational constant on earth and the distance moved by the object. Wnet=Fnet times Displacement.
![**Conservation of Energy Problem Explanation**
We are provided with a physics problem addressing a block interacting with two springs and involving an incline with friction. The information and equations used in the solution are outlined below.
**Problem Details:**
- A 4.6 kg block is initially at rest, compressed against a horizontal spring with a spring constant of \( k_1 = 2750 \, \text{N/m} \) and a compression distance of 0.40 m.
- After release, the block moves up a \( 28^\circ \) incline to a height of 2.8 m where it encounters a second spring with a spring constant of \( k_2 = 350 \, \text{N/m} \).
- The horizontal surfaces are frictionless, while the incline has a kinetic friction coefficient \( \mu_k = 0.16 \).
- The block’s final velocity when it reaches the top is 9.78 m/s.
- The task is to calculate the compression of the second spring when the block comes to a stop.
**Known Values:**
- \( k_1 = 2750 \, \text{N/m} \)
- \( k_2 = 350 \, \text{N/m} \)
- \( m = 4.6 \, \text{kg} \)
- \( W_{nc} = 0 \, \text{J} \)
- \( x_i = 0.40 \, \text{m} \)
- \( h_i = 0 \)
- \( v_i = 0 \, \text{m/s} \)
- \( h_f = 2.8 \, \text{m} \)
**Energy Equation Explanation:**
The conservation of mechanical energy is considered without non-conservative work (\( W_{nc} = 0 \)):
\[
W_{nc} + mgh_i + \frac{1}{2} k_1 x_i^2 + \frac{1}{2} mv_i^2 = mgh_f + \frac{1}{2} k_2 x_f^2 + \frac{1}{2} mv_f^2
\]
Simplifying for when the final velocity and non-conservative work are zero:
\[
\frac{1}{2} k_1 x_i^2 = mgh_f + \frac{1}{2}](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6c806aca-d4f4-488e-823b-e6a92c863c9f%2Fd6950d16-520e-4443-ae1e-fd293d4b9ce0%2Ff971czj_processed.png&w=3840&q=75)
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