Question

A 964-kg car starts from rest at the bottom of a drive way and has a speed of 3.00 m/s at a point where the drive way has risen a vertical height of 0.600 m. Friction and the drive force produced by the engine are the only two non conservative forces present. Friction does -2870 J of work. How much work does the engine do?

Answer 1

Answer:

Work done by the engine is 12876.32 joules.

Explanation:

We know that,

$\mathrm{W}=\mathrm{KE}+\mathrm{PE}+\mathrm{W}_{\mathrm{f}}$

Where,

W = work done by engine

$\mathrm{KE}=\mathrm{kinetic} \text { energy at top of driveway }=\frac{1}{2}\left(\mathrm{mV}^{2}\right)$

m = mass of the car = 964kg.

V = speed of car at top of driveway = 3 m/sec

PE = potential energy at top of driveway = m × g × h

$\mathrm{g}=\text { acceleration due to gravity }=9.8 \mathrm{m} / \mathrm{s}^{2}$

h = vertical height of driveway = 0.6 m

$\mathrm{W}_{\mathrm{t}}=\text { work done against friction }=2870$

Substituting values,

$W=\frac{1}{2} \times(964) \times\left(3^{2}\right)+964 \times(9.8) \times(0.6)+2870$

$W=0.5 \times 964 \times 9+964 \times 9.8 \times 0.6+2870$

$\mathrm{W}=4338+5668.32+2870$

W = 12876.32 joules  

Work done by the engine is 12876.32 joules.