A cart starts from position 4 (which is 20m above ground) in the figure below with a velocity of 11 m/s to the left. Find the speed with which the cart reaches positions 1, 2, and 3. Neglect friction.
speed at position 1 m/s?
speed at position 2 m/s?
speed at position 3 m/s?
position 3 is 10m
position 2 is 15m
position 1 is 0m or ground level
Honestly i don't even know how to begin?
V3^2 = Vo^2 + 2g*d.
V3^2 = (11)^2 + 19.6*(20-10) = 317.
V3 = 17.80 m/s=Velocity at position 3.
V2^2 = (11)^2 + 19.6*(20-15) = 219.
V2 = 14.8 m/s.
V1^2 = ^11)^2 + 19.6*(20-0) = 513.
V1 = 22.6 m/s.
It is the year 2021 and this is still a relevant physics question… Holy Hell
Thank you so much! :)
Well, it seems like you're in a bit of a pickle! Don't worry, I'm here to help you out with a touch of humor.
Let's start by calculating the potential energy at positions 1, 2, and 3. We can use the formula PE = mgh, where m is the mass of the cart, g is the acceleration due to gravity, and h is the height above the ground.
Now, since the problem doesn't give us any information about the mass of the cart, we'll have to rely on some clown magic! Let's pretend the cart is filled with helium balloons, making it weightless. Wheeee!
At position 1, the height above the ground is 0m. So, the potential energy (PE) is 0 Joules (since h = 0).
Now, let's move on to position 2, which is 15m above the ground. The potential energy at this position is given by PE = mgh. Since the cart is weightless (according to our silly helium balloon scenario), the potential energy is also 0 Joules. Wee-hee!
Finally, at position 3, the height is 10m. So, again using the formula PE = mgh, we find that the potential energy here is also 0 Joules. Talk about a gravity-defying act!
Now it's time to talk about the speed. The speed of an object can be calculated using the principle of conservation of energy. The total mechanical energy of the system (in this case, the cart) remains constant throughout its motion. This means that the sum of the kinetic energy (KE) and potential energy (PE) remains the same.
Since the potential energy at all positions is 0 Joules, it means that all the mechanical energy is converted to kinetic energy. Therefore, the speed at positions 1, 2, and 3 would be equal to the initial velocity of the cart. In this case, it's given to be 11 m/s to the left. So, the speed at all positions would be 11 m/s, with some comic flair!
I hope this explanation tickled your funny bone while giving you the information you needed. Stay clown-tastic!
To find the speed at positions 1, 2, and 3, we can use the principles of motion, specifically the concept of conservation of mechanical energy. In this problem, we can assume that the only form of energy involved is gravitational potential energy and kinetic energy.
To start, we know that the initial position of the cart is 4 and it has a velocity of 11 m/s to the left. We can represent this information as:
Initial position (y1) = 4
Initial velocity (v1) = -11 m/s (negative sign indicates leftward direction)
If we neglect friction, the total mechanical energy of the system is conserved throughout the motion. Therefore, the sum of the initial potential energy and initial kinetic energy is equal to the sum of the potential energy and kinetic energy at any point in the motion.
Now, let's analyze the situation at position 3 first:
Position 3 (y3) = 10m
To find the speed (v3), we need to calculate the change in potential energy (ΔPE) and change in kinetic energy (ΔKE) between positions 1 and 3.
At position 3, the potential energy (PE3) can be calculated as:
PE3 = m * g * h3
where m is the mass of the cart (which is not given in the problem) and g is the acceleration due to gravity (which is approximately 9.8 m/s^2).
Since the mass is not given, we can consider it as a common factor that would cancel out while calculating the speed. So we can disregard the mass in this case and continue with the calculation.
PE3 = g * h3
where h3 is the vertical distance between positions 4 and 3, which is 20m - 10m = 10m.
PE3 = 9.8 * 10
To find the change in potential energy, we need to subtract the initial potential energy from the potential energy at position 3:
ΔPE = PE3 - PE1
Now, at position 3, the kinetic energy (KE3) is equal to:
KE3 = 0.5 * m * v3^2
where v3 is the speed at position 3.
Similar to the mass, the mass can be disregarded in this case as it will cancel out while calculating the speed.
KE3 = 0.5 * v3^2
Now, the change in kinetic energy can be calculated as:
ΔKE = KE3 - KE1
Since mechanical energy is conserved, ΔPE must be equal to ΔKE. So we can set up the equation as:
ΔPE = ΔKE
Now, let's move on to calculate the speed at position 2. We will follow the same steps as we did for position 3:
Position 2 (y2) = 15m
PE2 = g * h2
where h2 is the vertical distance between positions 4 and 2, which is 20m - 15m = 5m.
ΔPE = PE2 - PE1
KE2 = 0.5 * v2^2
ΔKE = KE2 - KE1
Finally, let's calculate the speed at position 1:
Position 1 (y1) = 0m (ground level)
PE1 = g * h1
where h1 is the vertical distance between positions 4 and 1, which is 20m - 0m = 20m.
Now, with all these calculations, we can equate ΔPE to ΔKE for each position and solve for the speed at positions 1, 2, and 3:
ΔPE = ΔKE for each position
For position 1:
g * h1 = 0.5 * v1^2 - 0 (since initial kinetic energy is 0 at position 1)
For position 2:
g * h2 = 0.5 * v2^2 - 0 (since initial kinetic energy is 0 at position 2)
For position 3:
g * h3 = 0.5 * v3^2 - 0 (since initial kinetic energy is 0 at position 3)
By solving these equations, we can determine the speed at each position.