Question

plz slove this plzzzzzzzz​

Answer 1

Answer:

Given:

Velocity of car = 10 m/s (final)

Constant Acceleration = 2 m/s²

Mass = 200 kg

Constant Resistance = 50N

To find:

Power exerted by the engine.

Formulas used:

Power = Net Force × Velocity

Calculation:

Force exerted by engine = m × a

=> F = 200 × 2

=> F = 400 N

Net Force = F - (Regarding Force)

=> F" = 400 - 50

=> F" = 350 N

Power exerted by engine

= F" × velocity

= 350 × 10

= 3500 watts

= 3.5 Kilowatts

Answer 2

Answer:

• Power (P) of the Engine is 3.5 Kilo Watts

Given:

1. Mass of Vechile (M) = 200 Kg.
2. Acceleration of Vechile (a) = 2 m/s².
3. Retardating Force (f) = 50 N.
4. Velocity of Vechile (v) = 10 m/s.

Explanation:

$\rule{300}{1.5}$

Applying Newton's Second law of motion,

$\large\bigstar \; {\boxed{\tt F = Ma}}$

$\bold{Here}\begin{cases}\text{F Denotes Force} \\ \text{M Denotes Mass} \\ \text{a Denotes Acceleration}\end{cases}$

$\large{\boxed{\tt F = Ma}}$

Substituting the values,

$\large{\tt \hookrightarrow F = 200 \: Kg \times 2 \: m/s^2}$

$\large{\tt \hookrightarrow F = 200 \times 2}$

$\large{\tt \hookrightarrow F = 4 \times 100}$

$\large{\underline{\boxed{\tt F = 400 \: N}}}$

Now, Net Force,

$\large{\boxed{\tt F_{net} = F - f}}$

• F = Force Applied by vechile.
• f = Retarding Force.

$\large{\tt \hookrightarrow F_{net} = 400 \: N - 50 \; N}$

$\large{\tt \hookrightarrow F_{net} = 400 - 50}$

$\large{\underline{\boxed{\tt F_{net} = 350 \: N}}}$

$\rule{300}{1.5}$

$\rule{300}{1.5}$

We Know, From Instantaneous Power Formula,

$\large\bigstar \; {\boxed{\tt P = F.v}}$

$\bold{Here}\begin{cases}\text{F Denotes Force} \\ \text{P Denotes Power} \\ \text{v Denotes Velocity}\end{cases}$

$\large{\boxed{\tt P = F_{net}.v}}$

Substituting the values,

$\large{\tt \hookrightarrow P = F_{net} \times v}$

$\large{\tt \hookrightarrow P = 350 \: N \times 10 \: m/s}$

$\large{\tt \hookrightarrow P = 350 \times 10}$

$\large{\tt \hookrightarrow P = 3500}$

$\large{\tt \hookrightarrow P = 3.5 \times 10^3}$

$\huge{\boxed{\boxed{\tt P = 3.5 \: KW}}}$

∴ Power (P) of the Engine is 3.5 Kilo Watts.

$\rule{300}{1.5}$