# Question fcdef

Apr 27, 2015

The net force on level ground is 200 N, and the net force on an incline plane at a ${30}^{\text{o}}$ angle is -500 N.

On Level Ground

Calculate the acceleration using the equation $t = \frac{{v}_{f} - {v}_{i}}{a}$ .

a=(v_f-v_i)/t=(10"m/s"-0"m/s")/(5"s")=2 "m/s"^2

The net force causes acceleration. The net force is the vector sum of all of the forces acting on a body. In this case, the weight of the object is equal to but opposite the normal force. So they are balanced and their sum is $0 \text{N}$

The applied force and frictional force will determine the net force. Since no information about friction has been given, we can infer that the vector sum of the applied force and the frictional force is the net force.

${F}_{\text{net"=ma=100 "kg" xx 2"m/s"^2=200 "kg" * "m/s"^2=200 "N}}$

Frictionless Inclined Plane

Note: Because of the nature of an inclined plane, we will use the mass of the object, but not the acceleration used for level ground. Instead we will be using the acceleration due to gravity, $- 9.1 {\text{m/s}}^{2}$. On an inclined plane, $a = g \cdot \sin \theta$

We first need to determine the net force.

Weight = $m g$, where $g$ is the acceleration due to gravity, $- 9.81 {\text{m/s}}^{2}$. The weight vector, $m g$ is resolved into two components, one parallel to the inclined surface and the other perpendicular to the inclined surface. The parallel force vector has the equation mg*"sintheta#. The perpendicular force vector, $m g \cdot \cos \theta$ is perpendicular to the normal force, $\text{N}$, and they have equal magnitude but opposite directions, so their vector sum is $0 \text{N}$. There is no friction, so the parallel force vector gives us the net force.

Determine the weight, $m g$, of the $100 \text{kg}$ object by multiplying its mass times $- 9.81 {\text{m/s}}^{2}$.

$m g = 100 \text{kg" xx -9.81"m/s"^2=-981 "kg"*"m/s"^2=-981 "N}$

The net force is equal to the parallel force vector.

${F}_{\text{net"=mg*sintheta=-981 "N"*sin30^"o" = -490.5 "N"=-500 "N}}$

Let's go ahead and determine the acceleration of the object. As stated before, on an inclined plane, $a = g \cdot \sin \theta$.

$a = - 9.81 {\text{m/s"^2*sin30^"o"=-4.91"m/s}}^{2}$