Question

1. -/30 points a 0.36 - kg particle has a speed of 3.0 m/s at point a and kinetic energy of 8.0 j at point b. (a) what is its kinetic energy at a? j (b) what is its speed at point b? m/s (c) what is the total work done on the particle as it moves from a to b? j

Explanation:

Step1: Calculate kinetic energy at A

The formula for kinetic energy is $K = \frac{1}{2}mv^{2}$. Given $m = 0.36\ kg$ and $v = 3.0\ m/s$.
$K_A=\frac{1}{2}\times0.36\times(3.0)^{2}=1.62\ J$

Step2: Find speed at B

We know $K_B = 8.0\ J$ and $K=\frac{1}{2}mv^{2}$. Rearranging for $v$, we get $v=\sqrt{\frac{2K}{m}}$. Substituting $K = K_B=8.0\ J$ and $m = 0.36\ kg$, $v_B=\sqrt{\frac{2\times8.0}{0.36}}\approx6.67\ m/s$

Step3: Calculate work - done

By the work - energy theorem, $W=\Delta K=K_B - K_A$. Substituting $K_A = 1.62\ J$ and $K_B = 8.0\ J$, $W=8.0 - 1.62 = 6.38\ J$

Answer:

(a) $1.62\ J$
(b) $6.67\ m/s$
(c) $6.38\ J$