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?

Explanation:

Part 1: Free - Body Diagram (Visual Description)

• Forces in vertical direction:
• Gravitational force (\(F_g\)): Acts downward. The magnitude is calculated as \(F_g = mg\), where \(m = 25\space kg\) and \(g=9.8\space m/s^{2}\), so \(F_g=25\times9.8 = 245\space N\).
• Normal force (\(F_N\)): Acts upward, and in the vertical direction, since there is no acceleration in the vertical direction, \(F_N=F_g = 245\space N\).
• Forces in horizontal direction:
• Applied force (\(F_{app}\)): Acts to the right with a magnitude of \(95\space N\).
• Frictional force (\(F_f\)): Acts to the left (opposing the motion).

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, in the horizontal direction, \(F_{net}=F_{app}-F_f\).

Step 2: Calculate the net force first

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

Step 3: Solve for frictional force

From \(F_{net}=F_{app}-F_f\), we can re - arrange to \(F_f=F_{app}-F_{net}\). Substituting \(F_{app} = 95\space N\) and \(F_{net}=37.5\space N\), we have \(F_f=95 - 37.5 = 57.5\space N\).

Step 1: Apply Newton's second law

Newton's second law states that \(F_{net}=ma\), where \(m\) is the mass of the object and \(a\) is its acceleration.

Step 2: Substitute the values

Given \(m = 25\space kg\) and \(a=1.5\space m/s^{2}\), we calculate \(F_{net}=m\times a=25\times1.5\).

Step 3: Calculate the result

\(25\times1.5 = 37.5\space N\) (the net force acts in the direction of the applied force, i.e., to the right).

Answer:

The force of friction is \(\boldsymbol{57.5\space N}\) (acting to the left).

Part 3: Net force acting on the box