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
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
☆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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