# 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.

## Answer and Explanation:

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.