question on the bottom?????

Question

(a) What is the kenetic energy of a 2670-kg car that is traveling at 97.0 km/h?____kj
(b) How much work was fone to give the car this kenetic energy?____kj
(c) How much work was done to stop the car?_____kj

Answer 1

KE=1/2*m*v^2

convert km/h to m/s
since joules are N*m

97*1000/3600
=26.9 m/s

KE=2670*26.9
=71823

b abd c are both equal to a if we ignore friction

j

Answer 2

First lets convert the speed of the car:
v = 97 km/h * 1000 m/km / (3600 s/hour) = 26.94 m/s

a)
Kinetic energy is given by the equation

K = m v^2 / 2 = 2670 * 26.94^2 / 2 kg m^2 / s^s =
969 kJ.

b)
The work A is equal to the change of kinetic energy
when the car was accelerated from 0km/h (K0=0kJ)
to 97km/h (K1 = 969 kJ)

A = K1 - K0 = 969 kJ

c)
The work A, which was done to stop the car
is equal to the change of kinetic energy
when the car was decelerated from
97km/h (K1=969kJ) to 0km/h (K2 = 0kJ)

A = K2 - K1 = - 969 kJ