A 10 g particle has the potential energy shown in a graph with the y-axis label U(J) and the x axis is labeled in cm. There is a diagonal line from 4 J down to 1 cm and the line levels out on the x axis to 3 cm and rises again to a height of 2 J at 5 cm and levels out to 7 cm and rises to 6 J at 8 cm.
Part A: How much work does the force do as the particle moves from 2 cm to 6 cm?
W = ? J
Part B: What speed does the particle need at x = 2cm to arrive at x = 6cm with a speed of 10 m/s ?
v = ? m/s
I need a formula to try to find the W on question A
I tried Part b and got 26 and 9.6 and submitted but both answers were wrong! Got one more chance! Please help.
I just tried 2 J for A and missed it.
I left off the negative sign. -2 is correct.
In B Part I am using this formula:
first velocity = sq. rt of 2*(U + 1/2mv^2of final velocity /m
What do I use for U and can I use 10 g or should I convert this to kg in this formula? Does it matter that the x axis has cm and the final answer is m/s?
Since W = �ç F dx = area of the Fx-versus-x graph between xi and xf, the work done by the force as the
particle moves from i x = 2 cm to f x = 6 cm is −2 J.
The conservation of energy equation is f f i i . K +U = K +U We can see from the graph that i U = 0 J and
= 2 J in moving from x = 2 cm to x = 6 cm. The final speed is f v =10 m/s, so
1 2 1 2
2 2 i i 2 J + (0.010 kg)(10.0 m/s) = 0 J + (0.010 kg)v �Ë v = 22.4 m/s
sorry I copy and pasted and it came out weird.
Part a- take the area under the curve and multiply it by (-1) because W= -delta u (area under the curve).
part b- use Kf + Uf = Ki + Ui we know that Ui=0 and Uf= 2J (just by looking at the graph) they gave you your final velocity so we will solve for initial velocity by plugging in all the given information. Initial velocity is 22 m/s.
Part A: To find the work done as the particle moves from 2 cm to 6 cm, we need to calculate the area under the curve of the potential energy graph between these two points.
Since the graph consists of several line segments, we can break down the area into smaller sections and then sum them up.
First, calculate the area of the rectangle formed by the height of the potential energy (2 J) and the width of the displacement (4 cm - 2 cm = 2 cm):
Area = height x width
Area = 2 J x 2 cm
Area = 4 J.cm
Next, calculate the area of the triangle formed by the remaining displacement (6 cm - 4 cm = 2 cm) and the change in potential energy (2 J - 1 J = 1 J):
Area = (base x height) / 2
Area = (2 cm x 1 J) / 2
Area = 1 J.cm
Now, sum up the areas of the rectangle and the triangle to find the total work done:
Total Work = 4 J.cm + 1 J.cm
Total Work = 5 J.cm
Part B: To find the speed at x = 2 cm required to arrive at x = 6 cm with a speed of 10 m/s, we can use the principle of conservation of mechanical energy.
The mechanical energy at position x = 2 cm consists of the potential energy and the kinetic energy:
Mechanical Energy at x = 2 cm = Potential Energy + Kinetic Energy
At x = 2 cm, the potential energy is 2 J (read from the graph) and the kinetic energy is unknown (let's call it KE1).
At x = 6 cm, the potential energy is 6 J (read from the graph) and the kinetic energy is known (10 m/s).
Using the principle of conservation of mechanical energy, the total mechanical energy remains constant:
Potential Energy at x = 2 cm + KE1 = Potential Energy at x = 6 cm + 1/2 * mass * velocity^2
Since the mass is not given, we can assume it to be 10 g (0.01 kg). Plugging in the values:
2 J + KE1 = 6 J + 0.5 * 0.01 kg * (10 m/s)^2
2 J + KE1 = 6 J + 0.5 * 0.01 kg * 100 m^2/s^2
2 J + KE1 = 6 J + 0.5 J
KE1 = 6 J + 0.5 J - 2 J
KE1 = 4.5 J
Therefore, the particle must have a kinetic energy of 4.5 J at x = 2 cm to arrive at x = 6 cm with a speed of 10 m/s.
I hope that clears things up! Keep those math questions coming!
To find the work done as the particle moves from 2 cm to 6 cm, we can use the formula:
W = ∆U = U(final position) - U(initial position)
First, let's determine the initial and final potential energies. From the graph you provided, it can be seen that at 2 cm, the potential energy is 2 J, and at 6 cm, the potential energy is 6 J.
So, substituting these values into the formula:
W = 6 J - 2 J
W = 4 J
Therefore, the work done by the force as the particle moves from 2 cm to 6 cm is 4 Joules.
Now, let's move on to Part B - finding the speed of the particle at x = 2 cm in order to arrive at x = 6 cm with a speed of 10 m/s.
To solve this, we need to consider the conservation of mechanical energy. The total mechanical energy of the particle is the sum of its kinetic energy and potential energy, and it remains constant if no external forces (like friction or air resistance) are acting on the particle.
The formula for mechanical energy is:
E = KE + PE
Given that the particle has a potential energy of 2 J at x = 2 cm, we can calculate the kinetic energy at this position. Since we want to find the speed, we'll use the formula for kinetic energy:
KE = 1/2mv^2
Plugging in the known values:
2 J = 1/2 * 0.01 kg * v^2
Simplifying the equation, we find:
0.04 = v^2
v = √0.04
v = 0.2 m/s
Therefore, the particle needs to have a speed of 0.2 m/s at x = 2 cm in order to arrive at x = 6 cm with a speed of 10 m/s.
Please let me know if anything is unclear or if you need further assistance!