Question

A child's toy consists of a block that attaches to a table with a suction cup, a spring connected to that block, a ball, and a launching ramp. (Intro 1 figure) The spring has a spring constant k, the ball has a mass m, and the ramp rises a height y above the table, the surface of which is a height H above the floor.Initially, the spring rests at its equilibrium length. The spring then is compressed a distance s, where the ball is held at rest. The ball is then released, launching it up the ramp. When the ball leaves the launching ramp its velocity vector makes an angle theta with respect to the horizontal.Throughout this problem, ignore friction and air resistance. Part A Relative to the initial configuration (with the spring relaxed), when the spring has been compressed, the ball-spring system has gained kinetic energy gained potential energy lost kinetic energy lost potential energy Part B As the spring expands (after the ball is released) the ball-spring system gains kinetic energy and loses potential energy gains kinetic energy and gains potential energy loses kinetic energy and gains potential energy loses kinetic energy and loses potential energy Part C As the ball goes up the ramp, it gains kinetic energy and loses potential energy gains kinetic energy and gains potential energy loses kinetic energy and gains potential energy loses kinetic energy and loses potential energy Part D As the ball falls to the floor (after having reached its maximum height), it gains kinetic energy and loses potential energy gains kinetic energy and gains potential energy loses kinetic energy and gains potential energy loses kinetic energy and loses potential energy Part E Which of the graphs shown best represents the potential energy of the ball-spring system as a function of the ball's horizontal displacement? (Part E figure) Take the zero on the distance axis to represent the point at which the spring is fully compressed. Keep in mind that the ball is not attached to the spring, and neglect any recoil of the spring after the ball loses contact with it.Please Choose A B C D Part F Calculate vr, the speed of the ball when it leaves the launching ramp. Express the speed of the ball in terms of k, s, m, g, y, and/or H. vr =Part G With what speed will the ball hit the floor? Express the speed in terms of k, s, m, g, y, and/or H. vf =

A child's toy consists of a block that attaches to a table with a suction cup, a spring connected to that block, a ball, and a launching ramp. (Intro 1 figure) The spring has a spring constant , the ball has a mass , and the ramp rises a height above the table, the surface of which is a height above the floor.

Initially, the spring rests at its equilibrium length. The spring then is compressed a distance , where the ball is held at rest. The ball is then released, launching it up the ramp. When the ball leaves the launching ramp its velocity vector makes an angle with respect to the horizontal.

Throughout this problem, ignore friction and air resistance.

Part A

Relative to the initial configuration (with the spring relaxed), when the spring has been compressed, the ball-spring system has

gained kinetic energy gained potential energy lost kinetic energy lost potential energy

Part B

As the spring expands (after the ball is released) the ball-spring system

gains kinetic energy and loses potential energy gains kinetic energy and gains potential energy loses kinetic energy and gains potential energy loses kinetic energy and loses potential energy

Part C

As the ball goes up the ramp, it

gains kinetic energy and loses potential energy gains kinetic energy and gains potential energy loses kinetic energy and gains potential energy loses kinetic energy and loses potential energy

Part D

As the ball falls to the floor (after having reached its maximum height), it

gains kinetic energy and loses potential energy gains kinetic energy and gains potential energy loses kinetic energy and gains potential energy loses kinetic energy and loses potential energy

Part E

Which of the graphs shown best represents the potential energy of the ball-spring system as a function of the ball's horizontal displacement? (Part E figure) Take the "zero" on the distance axis to represent the point at which the spring is fully compressed. Keep in mind that the ball is not attached to the spring, and neglect any recoil of the spring after the ball loses contact with it.

Please Choose A B C D

Part F

Calculate $v_r$ , the speed of the ball when it leaves the launching ramp.

Express the speed of the ball in terms of , , , , , and/or .

$v_r$ =

Part G

With what speed will the ball hit the floor?

Express the speed in terms of , , , , , and/or .

$v_f$ =

Answer 1

Concepts and reason

The concepts required to solve the given problem are kinetic energy, potential energy of a spring, and law of conservation of energy.

First, calculate the gain or loss in the energy when the ball-spring system is either expands or relax. Later the speed of the ball can be calculated when the ball-spring system leaves the launching ramp and finally when ball hits the floor.

Fundamentals

Potential energy: It is defined as the stored energy by virtue of the position or configuration of a body.

Mathematically, potential energy can be expressed as,

$F\left( x \right) = \frac{{ - dU}}{{dx}}$

Here, $F\left( x \right)$ is the x component of net force acting on the particle and U is the potential energy as a function of the particle coordinate.

Kinetic energy: It is defined as the energy possessed due to its motion.

Mathematically, kinetic energy can be expressed as,

$K = \frac{1}{2}m{v^2}$

Here, K is the kinetic energy, m is the mass of the object, and v is the velocity of the object.

Law of conservation of energy: It states that the total mechanical energy of a system is conserved if the forces doing work on it are conservative.

$K + U = {\rm{constant}}$

Here, K is the kinetic energy and U is the potential energy.

Potential energy of a spring: It is expressed as,

$U\left( x \right) = \frac{1}{2}k{x^2}$

Here, k is the force constant and x is the extension in the spring.

(A)

The ball-spring system is at rest position. So, the kinetic energy of the spring is zero. When the spring is compressed, due to change in the shape of the spring, the spring possess the extra potential energy. Hence, the potential energy of the system increases.

Therefore, when the spring is compressed, the ball spring system has gained potential energy.

(B)

When the spring expands, the potential energy of the spring is converted into the kinetic energy of the ball. According to the law of conservation of energy, decrease in potential energy causes an increase in kinetic energy. Therefore, the system potential energy is loss, while the kinetic energy is gained.

(C)

According to the law of conservation of energy, when the ball goes up in upward direction, the kinetic energy of the ball is converted into potential energy of the ball. As the ball moves up, the ball loses kinetic energy and gains equal amount of potential energy.

(D)

As the ball reaches the maximum height, the ball has the maximum potential energy and its kinetic energy is zero. Again, when the ball reaches the ground from the maximum height because of law of conservation of energy, the maximum potential energy of the ball is converted into the kinetic energy of the ball.

(E)

Initially, the system has potential energy. When the ball moves up in the upward direction, as per the law of conservation of energy, the kinetic energy of the ball reduces and its potential energy increase. At the maximum height, the ball gains maximum potential energy and its kinetic energy is zero at maximum height. Similarly, when the ball reaches ground, the maximum potential energy at maximum height reduces to zero and the ball gains the maximum kinetic energy at the ground.

(F)

Using the law of conservation of energy,

${K_i} + {U_i} = {K_f} + {U_{\rm{f}}}$

Here, ${K_i}$ is the initial kinetic energy, ${U_i}$ is the initial potential energy, ${K_f}$ is the final kinetic energy, and ${U_{\rm{f}}}$ is the final potential energy

Substitute 0 J for K_i , $\frac{1}{2}k{s^2}$ for ${U_i}$ , $mgy$ for ${U_{\rm{f}}}$ , and $\frac{1}{2}mv_r^2$ for ${K_{\rm{f}}}$ in equation as follows:

$\begin{array}{c}\\{K_i} + {U_i} = {K_f} + {U_{\rm{f}}}\\\\0 + \frac{1}{2}k{s^2} = \frac{1}{2}mv_r^2 + mgy\\\\{v_r} = \sqrt {\left( {\frac{{k{s^2}}}{m} - 2gy} \right)} \\\end{array}$

(G)

The expression for law of conservation of energy is,

${K_i} + {U_i} = {K_f} + {U_{\rm{f}}}$

Here, ${K_i}$ is the initial kinetic energy, ${U_i}$ is the initial potential energy, ${K_f}$ is the final kinetic energy, and ${U_{\rm{f}}}$ is the final potential energy

Substitute 0 J for K_i, $\frac{1}{2}k{s^2}$ for ${U_i}$ , $mg\left( { - H} \right)$ for ${U_{\rm{f}}}$ , and $\frac{1}{2}mv_r^2$ for ${K_{\rm{f}}}$ in equation as follows:

$\begin{array}{c}\\{K_i} + {U_i} = {K_f} + {U_{\rm{f}}}\\\\0 + \frac{1}{2}k{s^2} = \frac{1}{2}mv_r^2 + mg\left( { - H} \right)\\\\{v_r} = \sqrt {\left( {\frac{{k{s^2}}}{m} + 2gH} \right)} \\\end{array}$

Here negative sign of height indicates that the height is in downward direction.

Ans: Part A

When the spring is compressed, the ball-spring system has gained potential energy.

Part B

As the spring expands, the ball spring system gains kinetic energy and loses potential energy. Option (a) is correct.

Part C

As the ball goes up the ramp, it loses kinetic energy and gains potential energy.

Part D

As the ball falls to the floor, it gains kinetic energy and loses potential energy.

Part E

The correct graph for the potential energy and distance is as follows:

Part F

The speed of the ball is ${v_r} = \sqrt {\left( {\frac{{k{s^2}}}{m} - 2gy} \right)}$ .

Part F

The speed of the ball is ${v_r} = \sqrt {\left( {\frac{{k{s^2}}}{m} + 2gH} \right)}$ .