Question

a 5.0-kg cart is moving horizontally at 6.0m/s. In order to change its speed to 10.0m/s, the net work done on the cart must be: 160J, 400J, 40J, 550J, 90J

Answer 1

Here we go ~

Formulae to be used here are :

`\quad\displaystyle \circ \: \rm kinetic \: \: energy = ( 1 )/(2) m {v}^(2)`

`\quad\displaystyle \circ \rm \: work \: done = change \: \: in \: \: KE`

The kinetic energy of cart while moving with velocity 6.0m/s can be calculated as :

`\qquad\displaystyle \tt \dashrightarrow \: (1)/(2) (5)(6) {}^(2)`

`\qquad\displaystyle \tt \dashrightarrow \: (1)/(2) (5)(36)`

Similarly, kinetic energy at velocity 10.0m/s would be :

`\qquad\displaystyle \tt \dashrightarrow \: (1)/(2) (5)(10 {}^{} ) {}^(2)`

`\qquad\displaystyle \tt \dashrightarrow \: (1)/(2) (5)(100)`

Next up ;

`\qquad\displaystyle \tt \dashrightarrow \: work \: done = KE_(final) - KE_(initial)`

`\qquad\displaystyle \tt \dashrightarrow \: WD = (1)/(2) (5)(100) - (1)/(2) (5)(36)`

`\qquad\displaystyle \tt \dashrightarrow \: WD = (1)/(2) (5)(100 - 36)`

`\qquad\displaystyle \tt \dashrightarrow \: WD = (1)/(2) (5)(64)`

`\qquad\displaystyle \tt \dashrightarrow \: WD = 5 * 32`

`\qquad\displaystyle \tt \dashrightarrow \: WD = 160 \: \: joules`

That's our required answer, n matches with choice A.) 160 J