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.

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 $a = 1.2 \text{ m/s}^2$, and assuming $g = 9.81 \text{ m/s}^2$.

Substitute the values:

Rounding to three significant figures:

4.2.2 Calculate the work done in lifting the crate.

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.

Rounding to three significant figures:

QUESTION 4.3: DYNAMICS

4.3.1 Calculate the maximum static frictional force 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 $g = 9.81 \text{ m/s}^2$.

Given coefficient of static friction $\mu_s = 0.35$.

Rounding to three significant figures:

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

From 4.3.1, $f_{s,max} \approx 77.8 \text{ N}$.
Since $F_{down} (103.6 \text{ N}) > f_{s,max} (77.8 \text{ N})$, 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.