# When a force is applied on a car of 100 kg by its engine; then its velocity increases from 54 km/h to 60 km/h; find the work done by the force. a. 200 kJ B. 20 kJ C. 2 kJ D. 120 kJ

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To solve, apply the work-energy theorem. The formula is:

`W=Delta KE = 1/2m(v_f^2-v_i^2)`

where

`W` - work done in Joules (J)

`Delta KE` - change in kinetic energy,

`m` - mass of object in kg,

`v_i` - initial velocity in m/s and

`v_f` - final velocity in m/s

So before subsitituting the values, convert the unit of velocities from km/h to m/s.

`v_i=54 (km)/hxx(1h)/(3600s)xx(1000m)/(1km)=15 m/s`

`v_f``=60(km)/hxx(1h)/(3600s)xx(1000m)/(1km)=16.67 m/s`

Then, plug-in m = 100, v_i = 15 and v_f=16.67 to the work-energy formula.

`W=1/2*100(16.67^2-15^2)=50(52.8889)=2644.445 J`

Then, divide it by 1000 to convert to kiloJoules, kJ.

`W=2644.445/1000=2.644445 kJ` **Among the given choices, the nearest answer is (C) 2kJ.**