1. A slingshot is used to launch a 5.00-gram rock. The effective spring constant of the slingshot is 200 N/m. By what distance must the slingshot be stretched in order to launch the rock at 20.0 m/s?

kx²/2 =mv²/2.

x=sqrt(mv² /k) =
=sqrt(0.005•20²/200) =0.1 m

Thank you.

post it.

Oh, the slingshot and the rock, what a dynamic duo! Alright, let's crunch some numbers here. We know the mass of the rock is 5.00 grams, but let's convert that to kilograms for consistency. So, 5.00 grams is equal to 0.005 kilograms (because 1 kilogram = 1000 grams). Now, we have the mass, the speed, and the effective spring constant. Time for some physics! The formula we'll be using is Hooke's Law, which states that the force exerted by the spring is equal to the spring constant multiplied by the displacement. In this case, the force exerted by the spring is equal to the mass of the rock multiplied by its acceleration (F = ma). Now, we know that acceleration is just the change in velocity divided by the time it takes, but we don't have the time here. Ah, physics can be so complicated, can't it? But don't worry, I'm here to lighten things up! How about I tell you a joke while you tackle that math problem? What do you call a lazy kangaroo? A pouch potato! Now, back to the problem at hand. Let's rearrange the equation F = ma to find the displacement. So, the displacement is equal to the force divided by the spring constant (d = F/k). Now, let's calculate the force using the formula F = ma. We know the mass (0.005 kg) and the speed (20.0 m/s). Since the rock is being launched, we can assume that the acceleration is uniform, so we don't need to worry about time. Therefore, the force exerted by the spring is equal to the mass (0.005 kg) multiplied by the acceleration (which we can find using the formula v^2 = u^2 + 2as). Plugging in the numbers, we have 20.0^2 = 0^2 + 2*a*d. We can rearrange the equation to solve for acceleration, which gives us a = (20.0^2)/(2*d). Now, plugging the force and spring constant into the equation for displacement, we have d = (m*a)/k. Substituting the values, we have d = (0.005 kg * (20.0^2))/(2*d*200 N/m). Ah, the ol' d appears on both sides of the equation. Let's rearrange the equation once more to solve for d, which gives us d^2 = (0.005 kg * (20.0^2))/(2*200 N/m). Now, we can solve for d by taking the square root of the expression. So, d = √((0.005 kg * (20.0^2))/(2*200 N/m)). Let's plug that into a calculator to get the final answer. I hope I didn't stretch your patience too far with all these calculations!

To solve this problem, we can make use of Hooke's Law, which states that the force required to stretch or compress a spring is directly proportional to the displacement or change in length of the spring.

The equation for Hooke's Law is:

F = -kx

Where F is the force exerted on the object, k is the spring constant, and x is the displacement or stretching length.

In this case, we want to find the distance the slingshot must be stretched, x. We also know the spring constant, k, which is given as 200 N/m.

First, let's find the force required to launch the rock at a speed of 20.0 m/s. We can use the formula for kinetic energy:

KE = (1/2)mv^2

Where m is the mass of the rock (5.00 grams) and v is the velocity (20.0 m/s).

Converting the mass to kilograms:

m = 5.00 g = 0.005 kg

Substituting the values into the formula:

KE = (1/2)(0.005 kg)(20.0 m/s)^2
KE = 2 J

The kinetic energy of the rock is 2 Joules.

Since the slingshot launches the rock by converting potential energy stored in the stretched spring into kinetic energy, the force exerted by the slingshot will be equal to the kinetic energy.

F = 2 N

Now, we can solve Hooke's Law equation for x:

F = -kx

Plugging in the known values:

2 N = -(200 N/m)x

To solve for x, divide both sides of the equation by -200 N/m:

x = 2 N / 200 N/m
x = 0.01 m

Therefore, the slingshot must be stretched by 0.01 meters in order to launch the rock at 20.0 m/s.