Question

1. -/3 points a 0.41 - kg particle has a speed of 5.0 m/s at point a and kinetic energy of 8.1 j at point b. (a) what is its kinetic energy at a? (b) what is its speed at point b? (c) what is the total work done on the particle as it moves from a to b?

Explanation:

Step1: Calculate kinetic energy at A

The formula for kinetic energy is $K = \frac{1}{2}mv^{2}$. Given $m = 0.41\ kg$ and $v = 5.0\ m/s$, we substitute these values: $K_A=\frac{1}{2}\times0.41\times(5.0)^{2}=\frac{1}{2}\times0.41\times25 = 5.125\ J$.

Step2: Calculate speed at B

We know $K_B = 8.1\ J$ and $K=\frac{1}{2}mv^{2}$. Rearranging for $v$, we get $v=\sqrt{\frac{2K}{m}}$. Substituting $K = 8.1\ J$ and $m = 0.41\ kg$, we have $v_B=\sqrt{\frac{2\times8.1}{0.41}}=\sqrt{\frac{16.2}{0.41}}\approx6.3\ m/s$.

Step3: Calculate total work - done

By the work - energy theorem, $W=\Delta K=K_B - K_A$. We found $K_A = 5.125\ J$ and $K_B = 8.1\ J$. So $W=8.1 - 5.125 = 2.975\ J$.

Answer:

(a) 5.125
(b) 6.3
(c) 2.975