Hector Arias

2023-03-30

A spring gun's spring has a constant force k =400 N/m and negligible mass. The spring is compressed 6.00 cm and a ball with mass 0.0300 kg is placed in the horizontal barrel against the compressed spring.The ball is then launched out of the gun's barrel after the spring is released. The barrel is 6.00 cm long, so the ball leaves the barrel at the same point that it loses contact with the spring. The gun is held so the barrel is horizontal. Calculate the speed with which the ballleaves the barrel if you can ignore friction. Calculate the speed of the ball as it leavesthe barrel if a constant resisting force of 6.00 Nacts on the ball as it moves along the barrel. For the situation in part (b), at what position along the barrel does the ball have the greatest speed?

?duairceasrxtg

Beginner2023-03-31Added 6 answers

We'll use the principle of conservation of mechanical energy. The total mechanical energy of the ball-spring system will be conserved throughout the motion, neglecting friction.

Let's begin by finding the potential energy stored in the compressed spring. The potential energy stored in a spring can be calculated using the formula:

$PE=\frac{1}{2}k{x}^{2}$

where $k$ is the spring constant and $x$ is the displacement from the equilibrium position.

Given:

Spring constant, $k=400\phantom{\rule{0.167em}{0ex}}\text{N/m}$

Compression, $x=6.00\phantom{\rule{0.167em}{0ex}}\text{cm}=0.06\phantom{\rule{0.167em}{0ex}}\text{m}$

Plugging in the values, we can calculate the potential energy stored in the spring:

$PE=\frac{1}{2}\times 400\phantom{\rule{0.167em}{0ex}}\text{N/m}\times (0.06\phantom{\rule{0.167em}{0ex}}\text{m}{)}^{2}$

Calculating this, we find:

$PE=0.72\phantom{\rule{0.167em}{0ex}}\text{J}$

Next, we'll determine the kinetic energy of the ball as it leaves the barrel. Since the barrel is horizontal, the potential energy stored in the spring will be fully converted into the kinetic energy of the ball.

The kinetic energy of an object can be calculated using the formula:

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

where $m$ is the mass of the object and $v$ is its velocity.

Given:

Mass of the ball, $m=0.0300\phantom{\rule{0.167em}{0ex}}\text{kg}$

Using the conservation of mechanical energy, we equate the potential energy stored in the spring to the kinetic energy of the ball:

$PE=KE$

$0.72\phantom{\rule{0.167em}{0ex}}\text{J}=\frac{1}{2}\times 0.0300\phantom{\rule{0.167em}{0ex}}\text{kg}\times {v}^{2}$

Simplifying the equation, we can solve for the velocity $v$:

${v}^{2}=\frac{2\times 0.72\phantom{\rule{0.167em}{0ex}}\text{J}}{0.0300\phantom{\rule{0.167em}{0ex}}\text{kg}}$

${v}^{2}=48\phantom{\rule{0.167em}{0ex}}{\text{m}}^{2}/{\text{s}}^{2}$

$v=\sqrt{48\phantom{\rule{0.167em}{0ex}}{\text{m}}^{2}/{\text{s}}^{2}}$

$v=6.93\phantom{\rule{0.167em}{0ex}}\text{m/s}$

So, the speed with which the ball leaves the barrel, neglecting friction, is 6.93 m/s.

Now, let's move on to part (b) of the problem, where a constant resisting force of 6.00 N acts on the ball as it moves along the barrel. In this case, we need to account for the work done by the resisting force.

The work done by a force is given by the formula:

$W=Fd$

where $W$ is the work done, $F$ is the force, and $d$ is the displacement.

Given:

Resisting force, $F=6.00\phantom{\rule{0.167em}{0ex}}\text{N}$

Barrel length, $d=6.00\phantom{\rule{0.167em}{0ex}}\text{cm}=0.06\phantom{\rule{0.167em}{0ex}}\text{m}$

Plugging in the values, we can calculate the work done by the resisting force:

$W=6.00\phantom{\rule{0.167em}{0ex}}\text{N}\times 0.06\phantom{\rule{0.167em}{0ex}}\text{m}$

$W=0.36\phantom{\rule{0.167em}{0ex}}\text{J}$

Since work is equal to the change in kinetic energy, we subtract the work done by the resisting force from the initial kinetic energy to find the final kinetic energy of the ball as it leaves the barrel.

Initial kinetic energy of the ball (from part (a)) is 0.72 J.

Final kinetic energy = Initial kinetic energy - Work done by resisting force

$K{E}_{\text{final}}=0.72\phantom{\rule{0.167em}{0ex}}\text{J}-0.36\phantom{\rule{0.167em}{0ex}}\text{J}$

$K{E}_{\text{final}}=0.36\phantom{\rule{0.167em}{0ex}}\text{J}$

Now, we can calculate the final velocity ${v}_{\text{final}}$ using the formula for kinetic energy:

$K{E}_{\text{final}}=\frac{1}{2}m{v}_{\text{final}}^{2}$

Plugging in the values, we can solve for ${v}_{\text{final}}$:

$0.36\phantom{\rule{0.167em}{0ex}}\text{J}=\frac{1}{2}\times 0.0300\phantom{\rule{0.167em}{0ex}}\text{kg}\times ({v}_{\text{final}}{)}^{2}$

$({v}_{\text{final}}{)}^{2}=\frac{2\times 0.36\phantom{\rule{0.167em}{0ex}}\text{J}}{0.0300\phantom{\rule{0.167em}{0ex}}\text{kg}}$

$({v}_{\text{final}}{)}^{2}=24\phantom{\rule{0.167em}{0ex}}{\text{m}}^{2}/{\text{s}}^{2}$

${v}_{\text{final}}=\sqrt{24\phantom{\rule{0.167em}{0ex}}{\text{m}}^{2}/{\text{s}}^{2}}$

${v}_{\text{final}}=4.90\phantom{\rule{0.167em}{0ex}}\text{m/s}$

So, the speed of the ball as it leaves the barrel with a constant resisting force of 6.00 N is 4.90 m/s.

Finally, to determine the position along the barrel where the ball has the greatest speed, we need to consider the work done by the resisting force. The work done is given by the formula:

$W=Fd$

Since the work done is constant, the magnitude of the force multiplied by the displacement will be the same at all positions along the barrel. Therefore, the ball will have the greatest speed at the position where it has traveled the greatest distance.

Since the barrel length is 6.00 cm, the ball will have the greatest speed when it reaches the end of the barrel, at a position of 6.00 cm from the starting point.

Thus, the ball has the greatest speed at the end of the barrel.

To summarize:

- The speed with which the ball leaves the barrel, neglecting friction, is 6.93 m/s.

- The speed of the ball as it leaves the barrel with a constant resisting force of 6.00 N is 4.90 m/s.

- The ball has the greatest speed at the end of the barrel (6.00 cm from the starting point).

Please let me know if anything is unclear or if you have any further questions!

Let's begin by finding the potential energy stored in the compressed spring. The potential energy stored in a spring can be calculated using the formula:

$PE=\frac{1}{2}k{x}^{2}$

where $k$ is the spring constant and $x$ is the displacement from the equilibrium position.

Given:

Spring constant, $k=400\phantom{\rule{0.167em}{0ex}}\text{N/m}$

Compression, $x=6.00\phantom{\rule{0.167em}{0ex}}\text{cm}=0.06\phantom{\rule{0.167em}{0ex}}\text{m}$

Plugging in the values, we can calculate the potential energy stored in the spring:

$PE=\frac{1}{2}\times 400\phantom{\rule{0.167em}{0ex}}\text{N/m}\times (0.06\phantom{\rule{0.167em}{0ex}}\text{m}{)}^{2}$

Calculating this, we find:

$PE=0.72\phantom{\rule{0.167em}{0ex}}\text{J}$

Next, we'll determine the kinetic energy of the ball as it leaves the barrel. Since the barrel is horizontal, the potential energy stored in the spring will be fully converted into the kinetic energy of the ball.

The kinetic energy of an object can be calculated using the formula:

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

where $m$ is the mass of the object and $v$ is its velocity.

Given:

Mass of the ball, $m=0.0300\phantom{\rule{0.167em}{0ex}}\text{kg}$

Using the conservation of mechanical energy, we equate the potential energy stored in the spring to the kinetic energy of the ball:

$PE=KE$

$0.72\phantom{\rule{0.167em}{0ex}}\text{J}=\frac{1}{2}\times 0.0300\phantom{\rule{0.167em}{0ex}}\text{kg}\times {v}^{2}$

Simplifying the equation, we can solve for the velocity $v$:

${v}^{2}=\frac{2\times 0.72\phantom{\rule{0.167em}{0ex}}\text{J}}{0.0300\phantom{\rule{0.167em}{0ex}}\text{kg}}$

${v}^{2}=48\phantom{\rule{0.167em}{0ex}}{\text{m}}^{2}/{\text{s}}^{2}$

$v=\sqrt{48\phantom{\rule{0.167em}{0ex}}{\text{m}}^{2}/{\text{s}}^{2}}$

$v=6.93\phantom{\rule{0.167em}{0ex}}\text{m/s}$

So, the speed with which the ball leaves the barrel, neglecting friction, is 6.93 m/s.

Now, let's move on to part (b) of the problem, where a constant resisting force of 6.00 N acts on the ball as it moves along the barrel. In this case, we need to account for the work done by the resisting force.

The work done by a force is given by the formula:

$W=Fd$

where $W$ is the work done, $F$ is the force, and $d$ is the displacement.

Given:

Resisting force, $F=6.00\phantom{\rule{0.167em}{0ex}}\text{N}$

Barrel length, $d=6.00\phantom{\rule{0.167em}{0ex}}\text{cm}=0.06\phantom{\rule{0.167em}{0ex}}\text{m}$

Plugging in the values, we can calculate the work done by the resisting force:

$W=6.00\phantom{\rule{0.167em}{0ex}}\text{N}\times 0.06\phantom{\rule{0.167em}{0ex}}\text{m}$

$W=0.36\phantom{\rule{0.167em}{0ex}}\text{J}$

Since work is equal to the change in kinetic energy, we subtract the work done by the resisting force from the initial kinetic energy to find the final kinetic energy of the ball as it leaves the barrel.

Initial kinetic energy of the ball (from part (a)) is 0.72 J.

Final kinetic energy = Initial kinetic energy - Work done by resisting force

$K{E}_{\text{final}}=0.72\phantom{\rule{0.167em}{0ex}}\text{J}-0.36\phantom{\rule{0.167em}{0ex}}\text{J}$

$K{E}_{\text{final}}=0.36\phantom{\rule{0.167em}{0ex}}\text{J}$

Now, we can calculate the final velocity ${v}_{\text{final}}$ using the formula for kinetic energy:

$K{E}_{\text{final}}=\frac{1}{2}m{v}_{\text{final}}^{2}$

Plugging in the values, we can solve for ${v}_{\text{final}}$:

$0.36\phantom{\rule{0.167em}{0ex}}\text{J}=\frac{1}{2}\times 0.0300\phantom{\rule{0.167em}{0ex}}\text{kg}\times ({v}_{\text{final}}{)}^{2}$

$({v}_{\text{final}}{)}^{2}=\frac{2\times 0.36\phantom{\rule{0.167em}{0ex}}\text{J}}{0.0300\phantom{\rule{0.167em}{0ex}}\text{kg}}$

$({v}_{\text{final}}{)}^{2}=24\phantom{\rule{0.167em}{0ex}}{\text{m}}^{2}/{\text{s}}^{2}$

${v}_{\text{final}}=\sqrt{24\phantom{\rule{0.167em}{0ex}}{\text{m}}^{2}/{\text{s}}^{2}}$

${v}_{\text{final}}=4.90\phantom{\rule{0.167em}{0ex}}\text{m/s}$

So, the speed of the ball as it leaves the barrel with a constant resisting force of 6.00 N is 4.90 m/s.

Finally, to determine the position along the barrel where the ball has the greatest speed, we need to consider the work done by the resisting force. The work done is given by the formula:

$W=Fd$

Since the work done is constant, the magnitude of the force multiplied by the displacement will be the same at all positions along the barrel. Therefore, the ball will have the greatest speed at the position where it has traveled the greatest distance.

Since the barrel length is 6.00 cm, the ball will have the greatest speed when it reaches the end of the barrel, at a position of 6.00 cm from the starting point.

Thus, the ball has the greatest speed at the end of the barrel.

To summarize:

- The speed with which the ball leaves the barrel, neglecting friction, is 6.93 m/s.

- The speed of the ball as it leaves the barrel with a constant resisting force of 6.00 N is 4.90 m/s.

- The ball has the greatest speed at the end of the barrel (6.00 cm from the starting point).

Please let me know if anything is unclear or if you have any further questions!

Three identical blocks connected by ideal strings are being pulled along a horizontal frictionless surface by a horizontal force $\overrightarrow{F}$ The magnitude of the tension in the string between blocks B and C is T=3.00N. Each block has mass m=0.400kg.

What is the magnitude F of the force?

What is the tension in the string between block A and block B??We have: A golf ball of mass 0.045 kg is hit off the tee at a speed of 45 m/s. The golf club was in contact with the ball for $5.0\times {10}^{-3}$ s. Find (a) the impulse imparted to the golf ball,and (b) the average force exerted on the ball by the golf club

???A box is sliding with a speed of 4.50 m/s on a horizontal surface when, at point P, it encounters a rough section. On the rough section, the coefficient of friction is not constant but starts at .100 at P and increases linerly with distance past P, reaching a value of .600 at 12.5 m past point P. (a) Use the work energy theorem to find how far this box slides before stopping. (b) What is the coefficient of friction at the stopping point? (c) How far would the box have slid iff the friciton coefficient didn't increase, but instead had the constant value of .1?

?

The spring in the figure (a) is compressed by length delta x . It launches the block across a frictionless surface with speed v0. The two springs in the figure (b) are identical to the spring of the figure (a). They are compressed by the same length delta x and used to launch the same block. What is the block's speed now?A small rock with mass 0.12 kg is fastened to a massless string with length 0.80 m to form a pendulum. The pendulum is swinging so as to make a maximum angle of 45 with the vertical. Air Resistance is negligible.

?

a) What is the speed of the rock when the string passesthrough the vertical position?

b) What is the tension in the string when it makes an angle of 45 with the vertical?

c) What is the tension in the string as it passes through the vertical?

Calculate the actual mechanical advantage of a lever.

The change in internal energy of a system that has absorbed 2 kcal of heat and does 500 J of work is

A) 6400J

B) 5400J

C) 7860J

D) 8900JThe electric field 10 cm from a long wire is 2.4 kN/C. If wire carries uniform charge, what will be the field strength at 40 cm from the wire?

$A)0.6\text{}kN/C;\phantom{\rule{0ex}{0ex}}B)150\text{}N/C;\phantom{\rule{0ex}{0ex}}C)75\text{}N/C;\phantom{\rule{0ex}{0ex}}D)4.8\text{}kN/C$Read each statement below carefully and state, with reasons, if it is true or false : (a) The net acceleration of a particle in circular motion is always along the radius of the circle towards the centre (b) The velocity vector of a particle at a point is always along the tangent to the path of the particle at that point (c) The acceleration vector of a particle in uniform circular motion averaged over one cycle is a null vector

Two paths lead to the top of a hill. One is shorter and steeper and the second one is longer but less steep. For which of the two paths is the gain in potential energy more?

A) Path I

B) Path II

C) Same for both the paths

D) Data insufficientWhat does negative $\mathrm{\u25b3}G$ mean?

What is a lever.

The gravitational unit of force is_?

What do you mean by $1\mathrm{eV}$?

Describe the difference between the coefficient static friction and kinetic friction by using examples.