The velocity of \(\displaystyle 12\frac{m}{s}\) can be broken into horizontal and vertical components by using trigonometry. Think of the figure below, where x and y velocity components of the total velocity are shown.

Use the total velocity, the x-component, and the y-component to form a right triangle below.

Treating \(\displaystyle 12\frac{m}{s}\) as the hypotenuse, x-component as the leg adjacent, and y-component as the leg opposite, you can conclude that the velocities are related through trigonometric identities.

\(\displaystyle v_y=v\sin\theta\)

\(\displaystyle v_x=v\cos\theta\)

Plugging in the given values, we can solve for the x and y velocity components.

\(\displaystyle v_y=(12\frac{m}{s})\sin(32^o)=6.36\frac{m}{s}\)

\(\displaystyle v_x=(12\frac{m}{s})\cos(32^o)=10.18\frac{m}{s}\)

First, find the net force by subtracting the opposing forces.

F = 1020N – 175N = 845 N to the west

Next, find acceleration using Newton's second law, \(\displaystyle F= ma\).

\(\displaystyle a = \frac{F}{m}\)

\(\displaystyle a = \frac{845 N}{15 kg} = 56 \frac{m}{s^2} (west)\)

The three blocks must remain in contact as they move, so they will each have the same velocity and acceleration regardless of their different masses. So, \(\displaystyle a_{A}=a_{B}=a_{C}\).

Know what forces are involved by drawing a force diagram.

The gravitational force is broken into the x and y components. The net force on the block in the y-direction is the normal force minus the y component of the gravitational acceleration (\(\displaystyle mgcos\theta\)).

\(\displaystyle F_N-mg\cos\theta=0\)

Notice that the net y-force is equal to zero to show that the block is not moving anywhere in the y-direction. Now, we can isolate the normal force.

\(\displaystyle F_N=mg\cos\theta\)

The net force in the x-direction is \(\displaystyle mg\sin\theta-F_{fr}=ma\). We know that the block is accelerating in the x-direction; therefore the net force is equal to \(\displaystyle ma\).

We can use the friction equations to substitute for the x-direction forces.

\(\displaystyle F_{fr}=\mu_k F_N=\mu_kmg\cos\theta\)

\(\displaystyle mg\sin\theta-\mu_kmg\cos\theta=ma\)

We can isolate the acceleration and solve using the provided values.

\(\displaystyle a=g\sin\theta-\mu_kg\cos\theta\)

\(\displaystyle a=(9.8\frac{m}{s^2})\sin(24^o)-(0.2)(9.8\frac{m}{s^2}\cos(24^o)\)

\(\displaystyle a=2.2\frac{m}{s^2}\)

The first step will be to find the final velocity of the car. We know the acceleration and time, so we can find the final velocity using kinematics. The initial velocity is zero, since the car starts at rest.

\(\displaystyle v_f=v_i+at\)

\(\displaystyle v=(0\frac{m}{s})+(2\frac{m}{s^2})(14s)=28\frac{m}{s}\)

Use this velocity and the mass of the car to solve for the final kinetic energy.

\(\displaystyle KE=\frac{1}{2}mv^2\)

\(\displaystyle KE=\frac{1}{2}(1000kg)(28\frac{m}{s})^2\)

\(\displaystyle KE=392,000J\)

We first need to find the net force acting on the ball during flight. We can then use the net force and Newton's second law to find the total acceleration on the ball.

\(\displaystyle F_{net}=F_g+F_{air}\)

\(\displaystyle F_{net}=mg+F_{air}\)

\(\displaystyle F_{net}=(2kg)(-9.8\frac{m}{s^2})+0.4N\)

\(\displaystyle F_{net}=-19.2N\)

Use this net force to find the acceleration.

\(\displaystyle F_{net}=ma\)

\(\displaystyle (-19.2N)=(2kg)a\)

\(\displaystyle a=-9.6\frac{m}{s^2}\)

From here, there are two ways to solve. One way uses kinematic equations, and the other uses energy. We will solve using energy.

Total energy must be conserved during the throw, so the initial kinetic energy must equal the final potential energy (since velocity is zero at the maximum height).

\(\displaystyle \frac{1}{2}mv_i^2+mgh_i=\frac{1}{2}mv_f^2+mgh_f\)

\(\displaystyle \frac{1}{2}mv_i^2+mg(0m)=\frac{1}{2}m(0\frac{m}{s})^2+mgh_f\)

\(\displaystyle \frac{1}{2}mv_i^2=mgh_f\)

Use the given initial velocity to find the final height.

\(\displaystyle \frac{1}{2}(2kg)(20\frac{m}{s})^2=(2kg)(9.6\frac{m}{s^2})(h)\)

\(\displaystyle h=\frac{\frac{1}{2}(2kg)(20\frac{m}{s})^2}{(2kg)(9.6\frac{m}{s^2})}\)

\(\displaystyle h=20.83m\)

The force that translates to the entire system is that of gravity acting on the mass hanging over the ledge.

\(\displaystyle F = mg = 14kg(10m/s^{2}) = 140 N\)

140N is the total force acting on the system, which has a mass equal to both blocks combined (65 kg + 14 kg = 79 kg). We can find the acceleration using Newton's second law.

\(\displaystyle a = \frac{F}{m}\)

\(\displaystyle a = \frac{140 N}{79 kg} = 1.8 m/s^{2}\)

We use Hooke's law equation to relate the force, displacement, and spring constant:

\(\displaystyle F_s=-kx\)

We are given the force and the displacement, allowing us to solve for the spring constant.

\(\displaystyle 100N=-k(-0.05m)\)

Note that the displacement is negative, since the spring is compressed. For springs, compressions represents a negative displacement, while stretching represents a positive displacement.

\(\displaystyle k=\frac{100N}{0.05m}=2000\frac{N}{m}\)

First, we need to solve for the spring constant by using the force on the spring. We can use Hooke's Law:

\(\displaystyle F=-kx\)

The magnitude of the force on the spring will be equal to the force of gravity on the mass, which is given to be \(\displaystyle 20N\). The distance the spring it stretched is in the downward direction, so we must use a negative displacement. Use these values to calculate the spring constant.

\(\displaystyle F=(20N)=-k(-0.10m)\)

\(\displaystyle k=200\frac{N}{m}\)

Next, use the spring energy equation with the displacement and spring constant to solve for the energy stored in the spring.

\(\displaystyle E=\frac{1}{2}kx^2\)

\(\displaystyle E=\frac{1}{2}(200\frac{N}{m})(-0.10m)^2\)

\(\displaystyle E=1J\)

We use the equation for centripetal force to find the radius:

\(\displaystyle F_c=\frac{mv^2}{r}\)

Since the string ties the ball to the axis, the force of tension will be equal to the centripetal force.

\(\displaystyle F_T=15N=\frac{mv^2}{r}\)

Use the given mass and velocity to solve for the radius.

\(\displaystyle 15N=\frac{(1kg)(3\frac{m}{s})^2}{r}\)

\(\displaystyle r=\frac{9\frac{m^2kg}{s^2}}{15N}=0.6m\)