Question

Mass- 800kgh-start point is at 50 m highpoint A is at 20m/s velocity and point B is at 12m in height

Answer 1

Given data

*The given mass is m = 800 kg

*The given height is h = 50 m

*The speed at point A is v_A = 20 m/s

*The height at point B is h_b = 12 m

(a)

The formula for the mechanical energy is given as

`ME=U_k+U_p`

*Here U_k = 0 J is the kinetic energy

*Here U_p = mgh is the potential energy

Substitute the known values in the above expression as

`\begin{gathered} ME=0+mgh \\ =800*9.8*50 \\ =392000\text{ J} \end{gathered}`

Hence, the mechanical energy is ME = 392000 J

(b)

The formula for the height at point A is given by the relation as

`\begin{gathered} ME=U_K+U_P \\ U_p=ME-U_k \end{gathered}`

Substitute the known values in the above expression as

`\begin{gathered} mgh_a=ME-(1)/(2)mv^2_A \\ 800*9.8* h_a=392000-(1)/(2)*800*(20)^2 \\ 7840h_a=232000 \\ h_a=29.5\text{ m} \\ \approx30\text{ m} \end{gathered}`

Hence, the height at point A is h_a = 30 m

(c)

The velocity at point B is calculated by the relation as

`\begin{gathered} ME=U_K+U_P \\ U_K=ME-U_P \end{gathered}`

Substitute the known values in the above expression as

`\begin{gathered} (1)/(2)mv^2_b=ME-mgh_b \\ (1)/(2)(800)(v^2_b)=392000-(800)(9.8)(12) \\ v_b=27.29\text{ m/s} \end{gathered}`