Question

1. of 3.2 m/s forward. determine the mass of the cart.
2. during a parachute jump, a 58 kg person opens the parachute and the total drag force acting on the person is 720 n up.

(a) calculate the net force acting on the person.
(b) determine the acceleration of the person.

1. a net force of magnitude 36 n gives an object of mass m₁ an acceleration of 6.0 m/s². the same net force gives m₁ and another object of mass m₂ fastened together an acceleration of 2.0 m/s². what acceleration will m₂ experience if the same net force acts on it alone?
2. a 1300 kg car accelerates at 1.6 m/s² e. a frictional force of 3800 n w is acting on the car.

(a) draw the fbd of the car.
(b) determine the applied force acting on the car.
table 1

Explanation:

6. (a)

Step1: Identify forces acting on the person

The force of gravity $F_g = mg$ acts down - ward and the drag force $F_d$ acts upward. The mass of the person $m = 58\ kg$ and $g=9.8\ m/s^{2}$. So $F_g=mg = 58\times9.8=568.4\ N$ (downward) and $F_d = 720\ N$ (upward).

Step2: Calculate the net - force

The net force $F_{net}=F_d - F_g$.
$F_{net}=720 - 568.4=151.6\ N$ (upward)

Step1: Use Newton's second law

Newton's second law is $F_{net}=ma$, where $F_{net}$ is the net force, $m$ is the mass and $a$ is the acceleration. We know $F_{net}=151.6\ N$ and $m = 58\ kg$.

Step2: Solve for acceleration

$a=\frac{F_{net}}{m}=\frac{151.6}{58}=2.61\ m/s^{2}$ (upward)

Step1: Find $m_1$ using Newton's second law

From $F = m_1a_1$, where $F = 36\ N$ and $a_1 = 6.0\ m/s^{2}$. Using $m=\frac{F}{a}$, we get $m_1=\frac{F}{a_1}=\frac{36}{6}=6\ kg$.

Step2: Find $m_1 + m_2$

Since $F=(m_1 + m_2)a_2$ and $F = 36\ N$, $a_2 = 2.0\ m/s^{2}$, then $m_1 + m_2=\frac{F}{a_2}=\frac{36}{2}=18\ kg$.

Step3: Find $m_2$

Since $m_1 + m_2=18\ kg$ and $m_1 = 6\ kg$, then $m_2=(m_1 + m_2)-m_1=18 - 6 = 12\ kg$.

Step4: Find the acceleration of $m_2$

Using $a=\frac{F}{m}$, with $F = 36\ N$ and $m = m_2=12\ kg$, we get $a=\frac{36}{12}=3.0\ m/s^{2}$

Answer:

$151.6\ N$ (upward)

6. (b)