Question

a vehicle starting from rest attains a speed of 72 kilometre per hour after covering a distance of 100 metres if the mass of the vehicle is 500 kilogram find the force exerted by the engine​

Answer 1

Answer:

1kN

Explanation:

72 kmph = 72 × 5/18 m/s = 20 m/s

a = (v² - u²) / (2S)

= [(20 m/s)² - 0] / (2 × 100 m)

= 2 m/s²

F = ma

= 500 kg × 2 m/s²

= 1000 N

= 1 kN

Force exerted by engine is 1 kN

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Answer 2

A vehicle starting from rest (means the initial velocity of the vehicle is 0 m/s) attains a speed of 72 km/hr after covering a distance of 100 m.

Also given that, the mass of the vehicle is 500 kg.

We have to find the force exerted by the engine.

From above data we have; initial velocity (u) is 0 m/s, final velocity (v) is 72 km/hr = 72*5/18 = 20 m/s and distance (s) is 100 m.

Using the Third Equation Of Motion:

v² - u² = 2as

Substitute the known values in the above formula,

→ (20)² - (0)² = 2(a)(100)

→ 400 - 0 = 200a

→ 400 = 200a

→ 2 = a

Therefore, the acceleration of the vehicle is 2 m/s².

Now, Force is defined as the product of mass and acceleration i.e.

F = ma

Substitute the value of F = 500 kg (given in question) and a = 2 m/s² (from above calculations)

→ F = 2 × 500

→ F = 1000 N or 1kN

Therefore, the force exerted by the vehicle is 1kN.