The velocity of  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  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.

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

Scalar quantities are those that can be described with magnitude only, as opposed to vectors, which include both magnitude and direction components. Distance, speed, and time are all scalars. Displacement is not a scalar, as it involves both the distance and the direction moved from a starting point. Velocity also includes a direction component, and is therefore a vector quantity.

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, .

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, .

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 ().

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.

The net force in the x-direction is . We know that the block is accelerating in the x-direction; therefore the net force is equal to .

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

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

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.

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

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.

Use this net force to find the acceleration.

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).

Use the given initial velocity to find the final height.

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



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.

First we can calculate the acceleration.

Using F = ma with the magnitude of the acceleration we can find the force.

To find the acceleration of the box traveling down the incline, the mass is not needed. Using the incline of the plane as the x-direction, we can see that there is no movement in the y-direction; therefore, we can use Newton's second, F = ma, in the x-direction.

There is only one force in the x-direction (gravity), however gravity is not just equal to “mg” in this case. Since the box is on an incline, the gravitational force will be equal to mgsin(30^o). Substituting force into F  =ma we find that mgsin(30^o) = ma. We can now cancel out masses and solve for acceleration.