Question

A net force of 5N is applied to a 13kg mass. what is its acceleration?​

Answer 1

Answer:

$\boxed {\boxed {\sf a \approx 0.4 \ m/s^2}}$

Explanation:

According to Newton's Second Law of Motion, force is the product of mass and acceleration.

$F= m \times a$

The net force is 5 Newtons and the mass is 13 kilograms. Let's convert the units for force to make the problem and unit cancellation easier.

• 1 Newton (N) is equal to 1 kilogram meter per square second (1 kg*m/s²)
• The net force of 5 N is equal to 5 kg*m/s²

Now we know the values for 2 variables:

• F= 5 kg*m/s²
• m= 13 kg

Substitute the values into the formula.

$5 \ kg *m/s^2 = 13 \ kg * a$

Since we are solving for the accleration we must isolate the variable, a. It is being multiplied by 13 kilograms and the inverse of multiplication is division. Divide both sides by 13 kg

$\frac {5 \ kg *m/s^2}{13 \ kg}= \frac{ 13 \ kg *a}{13 \ kg}$

$\frac {5 \ kg *m/s^2}{13 \ kg}=a$

The units of kilograms (kg) cancel.

$\frac {5 m/s^2}{13 }=a$

$0.384615385\ m/s^2=a$

The original measurements of force and mass ( 5 and 13) have 1 and 2 significant figures. We must round our answer to the least number of sig figs: 1.

For the number we found, that is the tenths place. The 8 in the hundredth place (0.384615385) tells us to round the 3 up to a 4.

$0.4 \ m/s^2 \approx a$

The acceleration is approximately 0.4 meters per square second.