Question

a 60 n force acts on a 15 kg object. find the acceleration.
equation: f = m × a
input values f 60 n × 15 kg 15²
solution + unit f = 450
part d: fill in the blanks (4 points — 1 point each)
1 as force increases, acceleration ______ when mass is constant.
2 as mass increases, acceleration ______ when force is constant.
3 the relationship between mass and acceleration is ______ proportional.
4 the relationship between force and acceleration is ______ proportional.
bonus (2 points)
explain, using newtons second law, why a small car accelerates faster than a large truck when both experience the same net force.

Explanation:

Step1: Recall Newton's second law

Newton's second law is \(F = ma\), where \(F\) is force, \(m\) is mass, and \(a\) is acceleration.

Step2: Solve for acceleration

We can rearrange the formula to \(a=\frac{F}{m}\). Given \(F = 60N\) and \(m = 15kg\), substitute the values: \(a=\frac{60}{15}\).

Step3: Calculate the value

\(\frac{60}{15}=4\).

Answer:

The acceleration is \(4m/s^{2}\).

For the fill - in - the - blanks:

1. According to \(F = ma\) (\(m\) constant), \(a=\frac{F}{m}\), so as \(F\) increases, \(a\) increases.
2. According to \(F = ma\) (\(F\) constant), \(a=\frac{F}{m}\), so as \(m\) increases, \(a\) decreases.
3. From \(a=\frac{F}{m}\) (\(F\) constant), when \(m\) increases, \(a\) decreases. So the relationship between \(m\) and \(a\) is inversely proportional.
4. From \(a=\frac{F}{m}\) (\(m\) constant), when \(F\) increases, \(a\) increases. So the relationship between \(F\) and \(a\) is directly proportional.

For the bonus:
According to \(F = ma\), \(a=\frac{F}{m}\). If \(F\) is the same, and the mass of a small car (\(m_{car}\)) is less than the mass of a large truck (\(m_{truck}\)) (\(m_{car}a_{truck}\) (because when the numerator \(F\) is the same and the denominator is smaller, the fraction is larger).