Question

c. determine the kinetic energy at b.
ke at b 700.12 joules

d. determine the speed at b
speed at b 6.41 m/s

e. determine the potential energy at c
pe at c 733.84 joules

f. determine the kinetic energy at c
ke at c 322.4 joules

g. determine the speed at c
speed at c 4.42 m/s

h. determine the speed at f
speed at f 7.92 m/s

Explanation:

Step1: Use energy conservation

Assume total mechanical energy is conserved: $E_{total} = KE_B + PE_B = KE_C + PE_C$. First find total energy from point B.
We know $KE_B = 700.12\ \text{J}$. To find $PE_B$, we can use point C's known $PE_C=733.84\ \text{J}$ and the speed at C ($v_C=4.42\ \text{m/s}$) to confirm mass first, or use total energy from point B.
First, calculate mass from point B: $KE_B = \frac{1}{2}mv_B^2$, so $m = \frac{2KE_B}{v_B^2}$
$m = \frac{2\times700.12}{(6.41)^2} = \frac{1400.24}{41.0881} \approx 34.08\ \text{kg}$

Step2: Calculate total energy

$E_{total} = KE_B + PE_B$. But we can also use point F: at F, speed is $7.92\ \text{m/s}$, assume PE at F is 0 (reference point), so $E_{total} = KE_F = \frac{1}{2}mv_F^2 = \frac{1}{2}\times34.08\times(7.92)^2$
$E_{total} = 17.04\times62.7264 \approx 1068.86\ \text{J}$

Step3: Solve for KE at C

$KE_C = E_{total} - PE_C$
$KE_C = 1068.86 - 733.84 = 335.02\ \text{J}$ (rounded to match precision)

Answer:

335.02 Joules