Question

A force of 1.50 N acts on a 0.20 kg trolley so as to accelerate it along an air track.

The track and force are horizontal and in line. How fast is the trolley going after acceleration from rest through 30 cm, if friction is negligible?

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Answer 1

Answer:

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Explanation:

The work done by the force causes, and is equal to, the increase in K.E. of the cart,

Therefore,

Work done=(K.E.)end - (K.E.)start

or Fs cos0°=½mvf2 - 0

substituting gives

(1.50N)(0.30m)=½(0.20kg)vf2

from which

vf=2.1m/s.

Answer 2

Given:

Force acting on the trolley, F = 1.5N

Mass of the trolley, m = 0.2Kg

Displacement,

$s = 30cm = 3 \times {10}^{ - 2}$

As the vehicle starts from rest,

Hence u = 0

To find:

Velocity of the trolley

Solution:

Applying the third equation of motion,

$force = mass \times accelaration \\ a = \frac{1.5}{0.2} = 7.5m {s}^{ - 2} \\ {v}^{2} = {u}^{2} + 2as \\ {v}^{2} = 0 + 2 \times 7.5 \times 3 \times {10}^{ - 2} \\ {v}^{2} = 4.5 \\ v = \sqrt{4.5} = 2.12m {s}^{ - 1}$

Hence, v = 2.12 m/s is the required answer.

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