Question

A 950 kg cart is pulled up to point A (140 m) where it is released from rest. What will the speed of the cart be when it reaches point B(95m)? Assume the total mechanical energy of the cart+ earth system is conserved

Answer 1

Due to the total mechanical energy is conserved, you have that the potential energy of the cart at 140 m must be equal to the sum of the potential energy and kinetic energy of the cart at 95m.

Then, you can write:

`\begin{gathered} K_A=K_B+U_B \\ m\cdot g\cdot h_A=(1)/(2)mv^2_B+m\cdot g\cdot h_B \end{gathered}`

where

hA = 140m

hB = 95 m

g = 9.8 m/s^2

m = 950 kg

cancel out the mass m in the previous equation, and solve for vB:

`\begin{gathered} v_B=2g\cdot h_A-2g\cdot h_B \\ v_B=\sqrt[]{2((9.8m)/(s^2))(140m)-2((9.8m)/(s^2))(95m)} \\ v_B=29.7(m)/(s) \end{gathered}`

Hence, when the cart reaches the point B, its speed is approximately 29.7 m/s