Question

a student is pushing a 90kg box across a floor by applying a 370n force to the right. the coefficient of static friction between the box and the floor is 0.4 and the coefficient of kinetic friction is 0.30. neglect drag. how many forces are acting on the box? what direction is the friction on the box? is the friction static or kinetic? what is the magnitude of frictional force on the box? what is the magnitude of the net force on the box? what is the magnitude of boxs acceleration?

Explanation:

Step1: Identify forces on the box

Forces on the box: gravitational force ($F_g = mg$), normal force ($F_N$), applied force ($F_{app} = 370N$ right), and frictional force ($F_f$). So vertical: $F_g$ (down) and $F_N$ (up); horizontal: $F_{app}$ (right) and $F_f$ (left). Total forces: 4.

Step2: Direction of friction

Friction opposes motion. Box is pushed right, so friction is left.

Step3: Static or kinetic friction

Box is moving (pushed across floor), so kinetic friction.

Step4: Magnitude of frictional force

First, find normal force. $F_g = mg = 90kg \times 9.8m/s^2 = 882N$. Since vertical forces balance, $F_N = F_g = 882N$. Kinetic friction: $F_f = \mu_k F_N = 0.30 \times 882N = 264.6N \approx 265N$.

Step5: Net force on the box

Net force $F_{net} = F_{app} - F_f = 370N - 264.6N = 105.4N \approx 105N$ (right).

Step6: Acceleration of the box

Using $F_{net} = ma$, $a = \frac{F_{net}}{m} = \frac{105.4N}{90kg} \approx 1.17m/s^2 \approx 1.2m/s^2$.

Answer:

1. Number of forces: 4
2. Friction direction: Left
3. Friction type: Kinetic
4. Frictional force: $\approx 265N$ (or $264.6N$)
5. Net force: $\approx 105N$ (or $105.4N$)
6. Acceleration: $\approx 1.2m/s^2$ (or $1.17m/s^2$)