Question

A 2000 kg truck traveling north at 34 km/h turns east and accelerates to 58 km/h. (a) What is the change in the truck's kinetic energy? (b) What is the magnitude of the change in its momentum?

Answer 1

Step-by-step explanation:

It is given that,

Mass of the truck, m = 2000 kg

Initial velocity of the truck, u = 34 km/h = 9.44 m/s

Final velocity of the truck, v = 58 km/h = 16.11 m/s

(a) Change in truck's kinetic energy,
`\Delta E=(1)/(2)m(v^2-u^2)`

`\Delta E=(1)/(2)* 2000\ kg* (16.11^2-9.44^2)`

`\Delta E=170418.5\ J`

`\Delta E=1.7* 10^5\ J`

(b) Change in momentum of the truck,
`\Delta p=m(v-u)`

`\Delta p=2000\ kg* (16.11-9.44)`

`\Delta p=13340\ kg-m/s`

Hence, this is the required solution.