A 2.38 10 3 kg car requires 5.4 kJ of work to move from rest to some final speed. During this...

Question:

A {eq}2.38 \times 10^3 {/eq} kg car requires 5.4 kJ of work to move from rest to some final speed. During this time, the car moves 28.1 m. Neglecting friction, find a) the final speed. Answer in units of m/s.

Work and Kinetic Energy

Assuming the energy is conserved at any point. The work done is supposed equal to the energy used by the object. The kinetic energy of an object is directly proportional to the product of the mass and the squared of the speed. This equation is given as {eq}E = \frac{1}{2}mV^2 {/eq}, where m and V are the mass and speed of the object.

Calculate the final velocity using the work-energy relation. That is

{eq}E = \frac{1}{2}mV^2 {/eq}

where

• {eq}E = 5.4\ kJ {/eq}
• {eq}m = 2.38 \times 10^3\ kg {/eq} is the mass
• {eq}V {/eq} is the speed

Putting V at one side of the equation and substituting all given values. Therefore,

{eq}V = \sqrt{\dfrac{2E}{m}}\\ V = \sqrt{\dfrac{2(5400\ J)}{2.38 \times 10^3\ kg}}\\ \boxed{V = 2.13\ m/s} {/eq}