# A cart starts from position 1 with a velocity of 17.0 m/s. Find the speed of the cart at...

## Question:

A cart starts from position 1 with a velocity of 17.0 m/s. Find the speed of the cart at positions 2 and 3 ignoring friction. Will the cart reach position 4?

position 2 = 12.0 m high

position 3 = 2.0 m high

position 4 = 20.m high

## Conservation of Energy:

The energy of an isolated object is conserved and remains constant at all time, until an external forces acts on it. For a car only possessing kinetic and potential energy, the magnitude of the change in the kinetic energy equals the magnitude of the change in potential energy at all times.

In our case, let the mass of the cart be m and let its potential energy at position 1 be zero. The cart has an initial kinetic energy of

{eq}K.E_i \ = \ 0.5 \ m \ \times \ (17.0 \ m/s)^2 \ = \ 144.5 \ m {/eq}

in Joules. Note that this is the total energy of the car and from conservation of energy, the total energy of the cart should always equal that amount.

{eq}E_{total} \ = \ 144.5 \ m {/eq}

in Joules. At position 2, the cart has a potential energy of

{eq}U_2 \ = \ m \ \times \ 9.81 \ m/s^2 \ \times \ 12.0 \ m \ = \ 117.72 \ m {/eq}

in Joules. Therefore, the kinetic energy of the cart at position 2 is equal to the total energy minus the potential energy at that point. This gives

{eq}0.5 \ m \ v^2 \ = \ 26.78 \ m {/eq}

where v is the speed at position 2. Solving for v, we get the speed of the cart at position 2 as

{eq}v \ = \ \sqrt{\dfrac{26.78}{0.5}} \ = \ \mathbf{7.32 \ m/s} {/eq}

correct to three significant figures.

Similarly for position 3, the kinetic energy of the cart is given by the total energy minus the potential energy at position 3.

{eq}\begin{align*} K.E_3 \ &= \ 144.5 \ m \ - \ (m \ \times \ 9.81 \ \times \ 2.0)\\ \\ 0.5 \ m \ v^2 \ &= \ 124.88 \ m\\ \\ \therefore \ v \ &= \ \sqrt{\dfrac{124.88}{0.5}} \\ \\ &= \ \mathbf{250 \ m/s} \end{align*} {/eq}

correct to three significant figures.

Note that the potential energy of the cart at position 4 is

{eq}U_4 \ = \ m \ \times \ 9.81 \ \times \ 20.0 \ = \ 196.2 \ m {/eq}

in Joules. This is greater than the total energy of the car, which is impossible since the total energy of the car has to be constant. Therefore, the cart {eq}\mathbf{cannot} {/eq} reach position 4. 