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?

Answer 1

Final velocity, V = 2.1213 m/s.

Step-by-step explanation:

Given the following data;

Force, F = 1.50N

Mass, m = 0.2kg

Displacement, S = 30cm to meters = 30/100 = 0.3m

To find the acceleration;

Force is given by the multiplication of mass and acceleration.

Mathematically, Force is given by the formula;

`F = ma` .......equation 1

Where;

• F represents force.

• m represents the mass of an object.

• a represents acceleration.

Making acceleration (a) the subject, we have;

`Acceleration (a) = (F)/(m)`

Substituting into the above equation, we have;

`Acceleration, a = (1.5)/(0.2)`

Acceleration, a = 7.5m/s²

Now to find the final velocity, we would use the third equation of motion;

`V^(2) = U^(2) + 2aS` .....equation 2

Where;

• V represents the final velocity measured in meter per seconds.

• U represents the initial velocity measured in meter per seconds.

• a represents acceleration measured in meters per seconds square.

• S represents the displacement measured in meters.

Note: Since the trolley is starting from rest, we know that it's initial velocity is 0.

Substituting the values into equation 2, we have;

`V^(2) = 0^(2) + 2*7.5*0.3`

`V^(2) = 0 + 4.5`

`V^(2) = 4.5`

Taking the square root, we have;

`V = \sqrt {4.5}`

V = 2.1213 m/s.

Therefore, the speed of the trolley after accelerating from rest through 0.3 m, if friction is negligible is 2.1213 m/s.