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.
4 answers
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?
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
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- 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.
2 answers