Question

What is the acceleration of a 113 kg object exposed to a 110 N force. (Round to one decimal place.)

Answer 1

Answer:

$\boxed {\boxed {\sf a \approx 1.0 \ m/s^2}}$

Explanation:

We are asked to find the acceleration of an object.

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

$F= m \times a$

We know the object has a mass of 113 kilograms and it is exposed to a 110 Newton force.

Let's convert the units of force to make the problem easier later. 1 Newton is equal to 1 kilogram meter sper second squared. So, the force of 110 Newtons is equal to 110 kilogram meters per second squared.

• F= 110 kg*m/s²
• m= 113 kg

Substitute the values into the formula.

$110 \ kg*m/s^2=113 \ kg * a$

We are solving for a, the acceleration, so we isolate this variable. It is being multiplied by 113 kilograms. The inverse operation of multiplication is division, so we divide both sides by 113 kg.

$\frac {110 \ kg*m/s^2 }{113 \ kg}= \frac{113 \ kg * a}{113 \ kg}$

$\frac {110 \ kg*m/s^2 }{113 \ kg}=a$

The units of kilograms (kg) cancel.

$\frac {110 m/s^2 }{113 }=a$

$0.973451327 \ m/s^2 = a$

We are asked to round to one decimal place or the tenths place. The 7 in the hundredths place tells us to round the 9 up to a 0, but then we must also round the 0 in the ones place to a 1.

$1.0 \ m/s^2 \approx a$

The acceleration of the object is approximately 1.0 meter per second squared.