Question

newton’s 2nd law – practice problems. name_______
date______
the following problems will require the use of
newton’s 2nd law. the formula states:
σf = m*a where σf is the net force or unbalanced force in (n)
m is the mass in (kg)
and a is the acceleration in (m/s²)

ex #1. determine the acceleration of a 5kg bowling ball when a 20n net force is
applied to the ball by a bowler.
work: ef = m * a
20n = 5 kg * a a = 20n / 5kg = 4 n/kg or 4 m/s²
5kg 5kg 5kg

ex #2. what net force is required to get an 12 kg object to accelerate at a rate
of 8m/s²?
work: ef = m * a
ef = 12kg 8 m/s² ef = 96 kg m/s² or 96n

practice problems: (show all work)

1. how much net force is required to get a 6kg package to accelerate 3m/s²?

1. determine the acceleration when a 50n net force is applied to a 7kg mass.

1. what acceleration would result when a 36n net force is applied to a 11kg?

1. what if the 36n net force is applied to a 4 kg mass?

1. how much acceleration would a -90n net force cause on a 18kg mass?

1. what net force had to act on a 3.5kg mass to cause an acceleration of

-2m/s/s?

1. if a mass accelerates 5 m/s² when a 10 n net force is applied, how large is

the mass?

Explanation:

Step1: Identify the formula

We use Newton's second law formula $\sum F = m \times a$. For problem 1, we need to find the net force ($\sum F$), given mass ($m = 6\,\text{kg}$) and acceleration ($a = 3\,\text{m/s}^2$).

Step2: Substitute values into the formula

Substitute $m = 6\,\text{kg}$ and $a = 3\,\text{m/s}^2$ into $\sum F = m \times a$. So $\sum F = 6\,\text{kg} \times 3\,\text{m/s}^2$.

Step3: Calculate the result

$6\times3 = 18$, so $\sum F = 18\,\text{kg}\cdot\text{m/s}^2$ which is $18\,\text{N}$ (since $1\,\text{N} = 1\,\text{kg}\cdot\text{m/s}^2$).

(For other problems, follow similar steps:

Problem 2:

Step1: Formula $\sum F = m \times a$, solve for $a$: $a=\frac{\sum F}{m}$

Step2: Substitute $\sum F = 50\,\text{N}$, $m = 7\,\text{kg}$: $a=\frac{50}{7}\approx7.14\,\text{m/s}^2$

Problem 3:

Step1: Formula $a=\frac{\sum F}{m}$

Step2: Substitute $\sum F = 36\,\text{N}$, $m = 11\,\text{kg}$: $a=\frac{36}{11}\approx3.27\,\text{m/s}^2$

Problem 4:

Step1: Formula $a=\frac{\sum F}{m}$

Step2: Substitute $\sum F = 36\,\text{N}$, $m = 4\,\text{kg}$: $a=\frac{36}{4}=9\,\text{m/s}^2$

Problem 5:

Step1: Formula $a=\frac{\sum F}{m}$

Step2: Substitute $\sum F = -90\,\text{N}$, $m = 18\,\text{kg}$: $a=\frac{-90}{18}=-5\,\text{m/s}^2$

Problem 6:

Step1: Formula $\sum F = m \times a$

Step2: Substitute $m = 3.5\,\text{kg}$, $a = -2\,\text{m/s}^2$: $\sum F = 3.5\times(-2)= -7\,\text{N}$

Problem 7:

Step1: Formula $m=\frac{\sum F}{a}$

Step2: Substitute $\sum F = 10\,\text{N}$, $a = 5\,\text{m/s}^2$: $m=\frac{10}{5}=2\,\text{kg}$)

Answer:

1. $\boldsymbol{18\,\text{N}}$
2. $\boldsymbol{\approx7.14\,\text{m/s}^2}$
3. $\boldsymbol{\approx3.27\,\text{m/s}^2}$
4. $\boldsymbol{9\,\text{m/s}^2}$
5. $\boldsymbol{-5\,\text{m/s}^2}$
6. $\boldsymbol{-7\,\text{N}}$
7. $\boldsymbol{2\,\text{kg}}$