A crate with a mass of 50 kg is pulled vertically upwards through a distance of 30 m with a constant acceleration of 1,2 m/s2. Calculate the tension in the rope. Calculate the work done in lifting the crate.

QUESTION 4.2: DYNAMICS

4.2.1 Calculate the tension in the rope.

Step 1: Apply Newton's second law of motion. The forces acting on the crate are the tension ($T$) upwards and its weight ($mg$) downwards. The net force causes an upward acceleration ($a$). Given mass $m = 50 \text{ kg}$, acceleration $a = 1.2 m/s^2$, and assuming $g = 9.81 m/s^2$. $\sum F = ma$ $T - mg = ma$ $T = mg + ma = m(g+a)$ Substitute the values: $T = 50 kg \times (9.81 m/s^2 + 1.2 m/s^2)$ $T = 50 kg \times 11.01 m/s^2$ $T = 550.5 \text{ N}$ Rounding to three significant figures: $\boxed{551 N}$

4.2.2 Calculate the work done in lifting the crate.

Step 1: Calculate the work done ($W$) by the tension force. The formula for work done is $W = F \times s \times \cos\theta$. Since the tension force and displacement are in the same direction ($\theta = 0^\circ$, so $\cos\theta = 1$), the formula simplifies to $W = T \times s$. Given distance $s = 30 \text{ m}$ and tension $T = 550.5 \text{ N}$ from 4.2.1. $W = 550.5 N \times 30 \text{ m}$ $W = 16515 \text{ J}$ Rounding to three significant figures: $\boxed{16500 J}$

QUESTION 4.3: DYNAMICS

4.3.1 Calculate the maximum static frictional force acting on the box.

Step 1: Calculate the normal force ($N$) acting on the box. The normal force is perpendicular to the inclined plane. The component of gravity perpendicular to the plane is $mg \cos \theta$. Given mass $m = 25 \text{ kg}$, angle $\theta = 25^\circ$, and assuming $g = 9.81 m/s^2$. $N = mg \cos \theta$ $N = 25 kg \times 9.81 m/s^2 \times \cos(25^\circ)$ $N = 245.25 N \times 0.9063077$ $N \approx 222.29 \text{ N}$

Step 2: Calculate the maximum static frictional force ($f_{s,max}$) using the formula $f_{s,max} = \mu_s N$. Given coefficient of static friction $\mu_s = 0.35$. $f_{s,max} = 0.35 \times 222.29 \text{ N}$ $f_{s,max} \approx 77.80 \text{ N}$ Rounding to three significant figures: $\boxed{77.8 N}$

4.3.2 Show that the box is indeed about to slide down the plane.

Step 1: Calculate the component of the gravitational force acting down the inclined plane. $F_{down} = mg \sin \theta$ $F_{down} = 25 kg \times 9.81 m/s^2 \times \sin(25^\circ)$ $F_{down} = 245.25 N \times 0.422618$ $F_{down} \approx 103.6 \text{ N}$

Step 2: Compare the force component down the plane ($F_{down}$) with the maximum static frictional force ($f_{s,max}$). From 4.3.1, $f_{s,max} \approx 77.8 \text{ N}$. Since $F_{down} (103.6 N) > f_{s,max} (77.8 N)$, the force pulling the box down the plane is greater than the maximum static friction that can oppose it. Therefore, the box will indeed slide down the plane.

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