5. The ball launcher in a pinball machine has a spring that has a force constant of 1.20 N/cm . The surface on which the ball moves is inclined 10.0° with respect to the horizontal. The spring is initially compressed 5.00 cm. Find (a) the launching speed of a 100-g ball when the plunger is released; (b) How far does the ball move along the incline? Friction and the mass of the plunger are negligible.
(a) Well, let's do some calculations to launch this ball into some pinball fun! The force constant of the spring is given as 1.20 N/cm, and the initial compression is 5.00 cm. Now, we can use Hooke's Law to find the force exerted by the spring: F = kx.
Here's the formula to calculate the launching speed (v) of the ball:
(v^2) = 2Fx/m,
where F is the force exerted by the spring, x is the initial compression, and m is the mass of the ball.
Given that the mass of the ball (m) is 100 g, which is equivalent to 0.1 kg, we can plug in the values.
F = (1.20 N/cm * 5.00 cm) = 6.00 N,
and m = 0.1 kg.
Now, let's find the launching speed:
v^2 = (2 * 6.00 N * 0.05 m) / 0.1 kg,
v^2 = 6.00 m^2/s^2,
v ≈ 2.45 m/s.
So, the launching speed of the ball is approximately 2.45 m/s.
(b) Now onto the distance the ball moves along the incline. We know the ball is moving up an incline of 10.0° with respect to the horizontal. We are in luck, as this can be conveniently solved using our old friend, trigonometry!
The distance the ball moves along the incline (s) can be found using the formula:
s = x / cos(θ),
where x is the initial compression and θ is the angle of the incline.
Given that x is 5.00 cm and θ is 10.0°, we can substitute these values:
s = 5.00 cm / cos(10.0°).
Now, let's solve that equation:
s ≈ 5.04 cm.
So, the ball moves approximately 5.04 cm along the incline.
Now, watch out for those bumpers and enjoy the game!
To find the launching speed of the ball, we can use the principle of conservation of mechanical energy. The potential energy stored in the compressed spring will be converted into kinetic energy when the spring is released.
Let's break down the problem into steps:
Step 1: Convert the force constant from cm to m
Given force constant: 1.20 N/cm
To convert it to N/m, we need to divide it by 100.
Force constant: 1.20 N/cm / 100 = 0.012 N/m
Step 2: Calculate the potential energy stored in the compressed spring
The potential energy stored in a spring can be calculated using the formula:
Potential Energy (PE) = 0.5 * k * x^2
where k is the force constant, and x is the compression of the spring.
Given compression: x = 5.00 cm = 0.05 m
Potential Energy (PE) = 0.5 * 0.012 N/m * (0.05 m)^2
Step 3: Convert the potential energy to kinetic energy
The potential energy stored in the spring will be converted into kinetic energy when the spring is released. The formula for kinetic energy is:
Kinetic Energy (KE) = 0.5 * m * v^2
where m is the mass of the ball, and v is the launching speed of the ball.
Given mass: m = 100 g = 0.1 kg
Potential Energy (PE) = Kinetic Energy (KE)
0.5 * 0.012 N/m * (0.05 m)^2 = 0.5 * 0.1 kg * v^2
Simplifying the equation:
0.006 * 0.0025 = 0.05 * v^2
0.000015 = 0.05 * v^2
Step 4: Solve for the launching speed of the ball
0.000015 = 0.05 * v^2
v^2 = 0.000015 / 0.05
v^2 = 0.0003
v = √0.0003
v ≈ 0.0173 m/s
Therefore, the launching speed of the ball is approximately 0.0173 m/s.
To find the distance the ball moves along the incline, we can use the formula for the distance traveled on an inclined plane:
Distance (d) = (v^2 * sin(2θ)) / g
where θ is the inclination angle and g is the acceleration due to gravity (9.8 m/s^2).
Given θ = 10.0° and g = 9.8 m/s^2:
Distance (d) = (0.0173 m/s^2 * sin(2 * 10.0°)) / 9.8 m/s^2
d = (0.0173 * sin(20°)) / 9.8
d ≈ 0.0021 m
Therefore, the ball moves approximately 0.0021 m along the incline.
To solve this problem, we need to use principles from Newton's laws of motion and energy conservation.
(a) To find the launching speed of the ball, we can use the principle of conservation of mechanical energy. The potential energy stored in the spring when it is compressed is equal to the kinetic energy of the ball when it is launched.
The potential energy stored in the spring can be calculated using the formula:
Potential Energy = 0.5 * k * x^2
where k is the force constant of the spring and x is the compression of the spring. In this case, k is given as 1.20 N/cm and x is given as 5.00 cm. However, we need to convert these values to SI units before plugging them into the formula.
1 cm = 0.01 m
1 N = 1 kg * m/s^2
So, the force constant in SI units is:
k = 1.20 N/cm * 0.01 m/cm = 0.012 N/m
And the compression in SI units is:
x = 5.00 cm * 0.01 m/cm = 0.05 m
Now we can calculate the potential energy:
Potential Energy = 0.5 * 0.012 N/m * (0.05 m)^2
Next, we equate the potential energy to the kinetic energy of the ball when it is launched:
Potential Energy = Kinetic Energy
0.5 * 0.012 N/m * (0.05 m)^2 = 0.5 * m * v^2
Simplifying, we get:
0.00003 N * m = 0.005 kg * v^2
Dividing both sides by 0.005 kg, we have:
v^2 = (0.00003 N * m) / (0.005 kg)
v^2 = 0.006 m^2/s^2
Taking the square root of both sides, we get:
v ≈ 0.077 m/s
Therefore, the launching speed of the ball is approximately 0.077 m/s.
(b) To find how far the ball moves along the incline, we can use the concept of work and energy. The work done by the force of gravity on the ball as it moves along the incline is equal to the change in potential energy.
The gravitational potential energy is given by:
Potential Energy = m * g * h
where m is the mass of the ball, g is the acceleration due to gravity, and h is the vertical height along the incline. In this case, the incline is at an angle of 10.0° with respect to the horizontal. So, the vertical height h can be calculated by multiplying the distance along the incline by the sin of the angle.
Now that we have all the necessary information, we can proceed with the calculations:
Mass of the ball, m = 100 g = 0.1 kg
Acceleration due to gravity, g = 9.8 m/s^2
Angle of the incline, θ = 10.0°
Vertical height, h = x * sin(θ)
where x is the distance along the incline.
We need to find the value of x, so let's rearrange the equation:
x = h / sin(θ)
Substituting the given values:
x = (0.05 m) / sin(10.0°)
Calculating x, we get:
x ≈ 0.289 m
Therefore, the ball moves approximately 0.289 meters along the incline.
a. Wb = m*g = 0.1kg * 9.8N/kg = 0.98 N. = Wt of the ball.
Fb = 0.98N @ 10o = Force of ball.
Fp = 0.98*sin10o = 0.170 N. = Force parallel to incline.
Fv = 0.98*cos10 = 0.965 N. = Force perpendicular to incline.
Fap = 1.2N/cm * (-5cm) = -6 N, = Force applied.
a=Fn/m=(Fap-Fp)/m
a=(-6+0.17)/0.1=-58.3m/s^2
Ekmax = Epmax.
0.5m*V^2 = mg*d.
0.5*0.1*V^2 = 0.98*5.
05*V^2 = = 4.9
V^2 = 98
V = 9.9 m/s.
b. V^2 = Vo^2 + 2a*d.
d = (V^2-Vo^2)/2a=(0-98)/-116.6=0.84 m.