Question

An engine of one metric ton is going up
an inclined plane, 1 in 2 at the rate of 36
kmph. If the coefficient of friction is
the power of engine is (in kilowatt)
13​

Answer 1

$\large\tt { P = mg \ sin \ θ + μmg \ cos \ θ)V}$

$\large\sf { = (1000 × 9.8 × \frac {1}{ \sqrt {5}} + \frac {1}{ \sqrt{3}} ×1000×9.8 × \frac {2}{ \sqrt{5}}}$

$\large\rm { 36 × \frac {5}{18} = 94.4 × 10³ W}$

$\large\rm { = 36 kilowatt }$

Answer 2

Answer:

P=mg sin θ+μmg cos θ)V

\large\sf { = (1000 × 9.8 × \frac {1}{ \sqrt {5}} + \frac {1}{ \sqrt{3}} ×1000×9.8 × \frac {2}{ \sqrt{5}}}=(1000×9.8×

5

1

+

3

1

×1000×9.8×

5

2

\large\rm { 36 × \frac {5}{18} = 94.4 × 10³ W}36×

18

5

=94.4×10³W

\large\rm { = 36 kilowatt }=36kilowatt