Question

energy essential version b
m=100kg
1
e_{gp}
e_k 11,250j
e_t
h50m
v
2
e_{gp}
e_k
e_t
h20.4m
v
3
e_{gp}
e_k
e_t
h
v26.6m/s

Explanation:

Step1: Calculate \( E_{gp} \) at position 1

Gravitational potential energy formula: \( E_{gp} = mgh \). Given \( m = 100\,\text{kg} \), \( g = 9.8\,\text{m/s}^2 \), \( h = 50\,\text{m} \).
\( E_{gp1} = 100 \times 9.8 \times 50 = 49000\,\text{J} \).
Total energy at position 1: \( E_{t1} = E_{gp1} + E_{k1} = 49000 + 11250 = 60250\,\text{J} \) (matches the given \( E_t \) if we consider possible rounding, but let's use conservation of energy: \( E_t \) is constant).

Step2: Calculate \( E_{gp} \) at position 2

\( h_2 = 20.4\,\text{m} \), so \( E_{gp2} = 100 \times 9.8 \times 20.4 = 100 \times 9.8 \times 20.4 = 19992\,\text{J} \approx 20000\,\text{J} \).
Since total energy \( E_t \) is conserved, \( E_{t2} = E_{t1} = 60250\,\text{J} \).
Kinetic energy at position 2: \( E_{k2} = E_{t2} - E_{gp2} = 60250 - 19992 = 40258\,\text{J} \approx 40260\,\text{J} \).
Velocity at position 2: \( E_k = \frac{1}{2}mv^2 \Rightarrow v = \sqrt{\frac{2E_k}{m}} = \sqrt{\frac{2 \times 40258}{100}} \approx \sqrt{805.16} \approx 28.37\,\text{m/s} \).

Step3: Calculate \( E_{gp} \) at position 3

Velocity at position 3: \( v_3 = 26.6\,\text{m/s} \). Kinetic energy: \( E_{k3} = \frac{1}{2} \times 100 \times (26.6)^2 = 50 \times 707.56 = 35378\,\text{J} \).
Total energy \( E_{t3} = E_{t1} = 60250\,\text{J} \), so \( E_{gp3} = E_{t3} - E_{k3} = 60250 - 35378 = 24872\,\text{J} \).
Height at position 3: \( h_3 = \frac{E_{gp3}}{mg} = \frac{24872}{100 \times 9.8} \approx 25.38\,\text{m} \approx 25.4\,\text{m} \).

(Note: Let's recheck with exact conservation. At position 1, \( E_{t1} = 49000 + 11250 = 60250\,\text{J} \).
Position 2:
\( E_{gp2} = 100 \times 9.8 \times 20.4 = 19992\,\text{J} \)
\( E_{k2} = 60250 - 19992 = 40258\,\text{J} \)
\( v_2 = \sqrt{\frac{2 \times 40258}{100}} = \sqrt{805.16} \approx 28.37\,\text{m/s} \)

Position 3:
\( E_{k3} = 0.5 \times 100 \times (26.6)^2 = 35378\,\text{J} \)
\( E_{gp3} = 60250 - 35378 = 24872\,\text{J} \)
\( h_3 = \frac{24872}{980} \approx 25.38\,\text{m} \))

Answer:

For position 1: \( E_{gp} = 49000\,\text{J} \), \( E_t = 60250\,\text{J} \), \( v \) (if needed) from \( E_k = 0.5mv^2 \), but \( E_k = 11250 \Rightarrow v = \sqrt{\frac{2 \times 11250}{100}} = \sqrt{225} = 15\,\text{m/s} \) (matches \( V \) if we see the first box's \( V \) is 15 m/s, likely a typo in the image, but let's use that).

For position 2:
\( E_{gp} \approx 20000\,\text{J} \), \( E_k \approx 40260\,\text{J} \), \( v \approx 28.4\,\text{m/s} \), \( E_t = 60250\,\text{J} \).

For position 3:
\( E_{gp} \approx 24870\,\text{J} \), \( E_k = 35378\,\text{J} \), \( h \approx 25.4\,\text{m} \), \( E_t = 60250\,\text{J} \).

(If the problem is to fill the boxes, here's the breakdown using conservation of energy \( E_t = \text{constant} = 60250\,\text{J} \)):

• Position 1: \( E_{gp} = 49000\,\text{J} \), \( E_k = 11250\,\text{J} \), \( E_t = 60250\,\text{J} \), \( v = 15\,\text{m/s} \) (since \( 0.5 \times 100 \times 15^2 = 11250 \)).

• Position 2:

\( E_{gp} = 100 \times 9.8 \times 20.4 = 19992\,\text{J} \approx 20000\,\text{J} \),
\( E_k = 60250 - 19992 = 40258\,\text{J} \approx 40260\,\text{J} \),
\( v = \sqrt{\frac{2 \times 40258}{100}} \approx 28.4\,\text{m/s} \),
\( E_t = 60250\,\text{J} \).

• Position 3:

\( v = 26.6\,\text{m/s} \), so \( E_k = 0.5 \times 100 \times 26.6^2 = 35378\,\text{J} \),
\( E_{gp} = 60250 - 35378 = 24872\,\text{J} \),
\( h = \frac{24872}{100 \times 9.8} \approx 25.4\,\text{m} \),
\( E_t = 60250\,\text{J} \).

(The key is using \( E_{gp} = mgh \), \( E_k = 0.5mv^2 \), and \( E_t = E_{gp} + E_k \) (conservation of energy, assuming no friction).)