Question

what is the work to be done to increase the velocity of a car from 30 km h-1 to 60 km h^-1 if the mass of the car us 1500 kg?​

Answer 1

Answer:

156375 J

Solution:-

Mass of the car=1500 kg

Intial velocity of car=30 km h^-1

$= \frac{30 \times 1000 \: m}{60 \times 60 \: sec}$

$= 8.33 \: m \: {s}^{ - 1}$

Similarly;the final velocity of the car

$v = 60 \: km \: {s}^{ - 1}$

$= 16.67 \: m \: {s}^{ - 1}$

The Intial kinetic energy of the car

$e_{ki} = \frac{1}{2}m \: {u}^{2}$

$= \frac{1}{2} \times 1500 \: kg \times {(8.33m \: {s}^{ - 1} }^{2})$

$= 52041.68 \: j$

The final kinetic energy of the car

$e_{kf} = \frac{1}{2} \times \frac{1}{2} \times {(16.67m \: {s}^{ - 1} })^{2}$

$= 208416.68 \: j$

_____________________

Thus the work done=change in kinetic energy

$= e_{kf} - e_{ki}$

$= 208416.68 - 52041.68$

=156375 J

Answer 2

Explanation:

Mass of the car=1500 kg

Intial velocity of car=30 km h^-1

= \frac{30 \times 1000 \: m}{60 \times 60 \: sec}=

60×60sec

30×1000m

= 8.33 \: m \: {s}^{ - 1}=8.33ms

−1

Similarly;the final velocity of the car

v = 60 \: km \: {s}^{ - 1}v=60kms

−1

= 16.67 \: m \: {s}^{ - 1}=16.67ms

−1

The Intial kinetic energy of the car

e_{ki} = \frac{1}{2}m \: {u}^{2}e

ki

=

2

1

mu

2

= \frac{1}{2} \times 1500 \: kg \times {(8.33m \: {s}^{ - 1} }^{2})=

2

1

×1500kg×(8.33ms

−1

2

)

= 52041.68 \: j=52041.68j

The final kinetic energy of the car

e_{kf} = \frac{1}{2} \times \frac{1}{2} \times {(16.67m \: {s}^{ - 1} })^{2}e

kf

=

2

1

×

2

1

×(16.67ms

−1

)

2

= 208416.68 \: j=208416.68j

_____________________

Thus the work done=change in kinetic energy

= e_{kf} - e_{ki}=e

kf

−e

ki

= 208416.68 - 52041.68=208416.68−52041.68

=156375 J