Physics, asked by liwehfqa, 4 months ago

A 3 000 kg car moving at a velocity of 30 km/h along SLEX is accelerated by a force of 2 600 N. What will be its velocity after 15 seconds? (neglect friction)

Answers

Answered by Ekaro
12

Given :

Mass of the car = 3000kg

Initial velocity = 30km/hr

Applied force = 2600N

Time interval = 15s

To Find :

Final velocity of car after 15s.

Solution :

❒ As per newton's second law of motion, force is defined as the rate of change of linear momentum.

Mathematically, F = dp / dt

  • Force is a polar vector quantity as it has a point of application.
  • SI unit : N

We know that, momentum is measured as the product of mass and velocity.

So we can say that, F = m(v - u) / dt

» m denotes mass

» v denotes final velocity

» u denotes initial velocity

» dt denotes time

Conversion :

we know that, 1 km/hr = 5/18 m/s

∴ 30 km/hr = 30 × 5/18 = 8.33 m/s

By substituting the given values;

➙ F = m(v - u) / dt

➙ 2600 = 3000(v - 8.33) / 15

➙ 2600 × 15/3000 = v - 8.33

➙ 13 = v - 8.33

➙ v = 13 + 8.33

v = 21.33 m/s

Answered by Anonymous
4

☆Answer☆

Mass of the car = 3000 kg

Initial velocity = 30 km/hr

Force = 2600 N

Time interval = 15 s

From Newton's 2nd law of motion:

F = dp/t

We have, momentum = mass×velocity

Then,

F = m(v-u)/t

changing In m/s

1 km/hr = 5/18 m/s

30 km/hr = 30 × 5/18 = 8.33 m/s

Substituting the values in formula, we get:

F = m(v-u) /t

2600 = 3000(v - 8.33)/15

2600 × 15/3000 = v-8.33

13 = v-8.33

v = 13+8.33

v = 21.33 m/s

Hence, the velocity of car after 15s is 21.33 m/s.

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