Question

question 6
a 25 kg box is sliding across a surface while experiencing a rightward applied force of 95 n. the box is accelerating at 1.5 m/s².
draw a free - body diagram showing all forces acting on the box.
calculate the force of friction acting on the box.
what is the net force acting on the box?
question 7
a 35 kg box is sliding across a surface at a constant speed while experiencing a leftward applied force of 120 n.
draw a free - body diagram showing all forces acting on the box.
calculate the force of friction acting on the box.
what is the net force acting on the box?

Explanation:

QUESTION 6

Part 1: Free - Body Diagram

The forces acting on the box are:

• Applied Force (\(F_{applied}\)): A right - ward force of 95 N.
• Frictional Force (\(F_{friction}\)): A left - ward force (opposing the motion).
• Gravitational Force (\(F_{gravity}\)): A downward force, \(F_{g}=mg\), where \(m = 25\space kg\) and \(g=9.8\space m/s^{2}\), so \(F_{g}=25\times9.8 = 245\space N\).
• Normal Force (\(F_{normal}\)): An upward force, equal in magnitude to the gravitational force (since there is no acceleration in the vertical direction), so \(F_{normal}=245\space N\).

To draw the free - body diagram:

• Draw a square to represent the box.
• Draw an arrow to the right for the applied force (length proportional to 95 N).
• Draw an arrow to the left for the frictional force (length to be determined from calculations).
• Draw an arrow downward for the gravitational force (length proportional to 245 N).
• Draw an arrow upward for the normal force (length proportional to 245 N).

Part 2: Calculate the force of friction

Step 1: Recall Newton's second law

Newton's second law is given by \(F_{net}=ma\), where \(F_{net}\) is the net force, \(m\) is the mass, and \(a\) is the acceleration. Also, the net force in the horizontal direction is \(F_{net}=F_{applied}-F_{friction}\) (since the applied force is to the right and friction is to the left).

Step 2: Calculate the net force first

We know that \(m = 25\space kg\) and \(a=1.5\space m/s^{2}\). Using \(F_{net}=ma\), we get \(F_{net}=25\times1.5=37.5\space N\) (to the right).

Step 3: Solve for the frictional force

From \(F_{net}=F_{applied}-F_{friction}\), we can re - arrange the formula to solve for \(F_{friction}\): \(F_{friction}=F_{applied}-F_{net}\). Substituting \(F_{applied} = 95\space N\) and \(F_{net}=37.5\space N\), we get \(F_{friction}=95 - 37.5=57.5\space N\) (to the left).

Step 1: Use Newton's second law

Newton's second law states that \(F_{net}=ma\). We know the mass \(m = 25\space kg\) and the acceleration \(a = 1.5\space m/s^{2}\).

Step 2: Calculate the net force

Substitute the values into the formula: \(F_{net}=25\times1.5 = 37.5\space N\) (to the right, since the box is accelerating to the right).

Step 1: Analyze the motion

The box is moving at a constant speed, which means the acceleration \(a = 0\space m/s^{2}\). According to Newton's second law \(F_{net}=ma\), so the net force \(F_{net}=0\) (since \(a = 0\)).

Step 2: Analyze the horizontal forces

In the horizontal direction, the net force \(F_{net}=F_{applied}-F_{friction}\) (taking left as positive). Since \(F_{net}=0\), we have \(F_{applied}-F_{friction}=0\).

Step 3: Solve for the frictional force

We know that \(F_{applied}=120\space N\) (left - ward). From \(F_{applied}-F_{friction}=0\), we get \(F_{friction}=F_{applied}=120\space N\) (right - ward).

Answer:

The force of friction is \(57.5\space N\) (left - ward).

Part 3: Net force acting on the box