Work

A warehouse worker pushes a pallet with a constant horizontal force of $250\ \text{N}$, moving it $12\ \text{m}$ along the floor in a straight line.

1. a) What angle does the force make with the displacement?
2. b) Calculate the work done by the worker.

A person walks their dog, pulling on the lead with a constant force of $80\ \text{N}$ at an angle of $60^\circ$ to the horizontal over a straight section of $25\ \text{m}$ ($\cos 60^\circ = 0,5$), as shown in the figure.

Calculate the work done.

In the laboratory, a trolley on an air track moves $\vec{d} = (5, 0)\ \text{m}$ while acted upon by a constant force $\vec{F} = (6, 8)\ \text{N}$, as shown in the figure.

1. a) Calculate the work using the dot product $W = \vec{F} \cdot \vec{d} = F_x \cdot d_x + F_y \cdot d_y$.
2. b) Explain why the vertical component of the force does not contribute to the work.

A constant force $\vec{F} = (15, -8)\ \text{N}$ acts on a machine component while the component moves through a displacement of $\vec{d} = (4, 2)\ \text{m}$, as shown in the figure.

1. a) Calculate the work using the dot product $W = \vec{F} \cdot \vec{d} = F_x \cdot d_x + F_y \cdot d_y$.
2. b) Check that the component $F_y$ does negative work and interpret what that sign means physically.

To load a fridge of $60\ \text{kg}$ into a van, it is pushed up a ramp of $3\ \text{m}$ at an angle of $20^\circ$ to the horizontal ($\sen 20^\circ \approx 0,34$; take $g = 10\ \text{m/s}^2$), as shown in the figure.

1. a) Calculate the component of the weight parallel to the ramp.
2. b) Calculate the work done by the weight during the ascent.
3. c) What minimum work must the person pushing do against that component of weight?

A box of $25\ \text{kg}$ starts from rest on the horizontal floor of a warehouse, where the coefficient of kinetic friction is $\mu = 0,2$. It is pushed with a constant horizontal force of $100\ \text{N}$ over $12\ \text{m}$ (take $g = 10\ \text{m/s}^2$), as shown in the figure.

1. a) Calculate the kinetic friction force, $F_{roz} = \mu \cdot N$, taking into account that on a horizontal floor the normal force is $N = m \cdot g$.
2. b) Calculate the work done by each of the forces (push, friction, weight, and normal force) and the net work.
3. c) Apply the work-energy theorem to find the speed of the box at the end of the $12\ \text{m}$.

The work-energy theorem states that the net work done on a body is equal to the change in its kinetic energy: $W_{neto} = \Delta E_c = \frac{1}{2} \cdot m \cdot v^2 - \frac{1}{2} \cdot m \cdot v_0^2$. A demonstration ball of $2\ \text{kg}$ increases its speed from $3\ \text{m/s}$ to $5\ \text{m/s}$ on an air track.

1. a) Calculate its initial kinetic energy and its final kinetic energy.
2. b) What is the net work done on the ball?

A car of $1000\ \text{kg}$ travelling at $20\ \text{m/s}$ brakes to a halt over a distance of $50\ \text{m}$ along a horizontal section.

1. a) Calculate the change in kinetic energy during braking.
2. b) Applying the work-energy theorem, find the work done by the net braking force.
3. c) What is the average value of this net braking force? State its sense.

On an ice rink, a puck of mass $m$ is launched with an initial speed of $v_0 = 10\ \text{m/s}$, and slides in a straight line until coming to a stop due to friction ($\mu = 0,25$; take $g = 10\ \text{m/s}^2$).

1. a) Write the expression for the work done by friction and, applying the work-energy theorem, obtain the stopping distance $d$ as a function of $v_0$, $\mu$, and $g$.
2. b) Calculate this distance numerically.
3. c) Reason, from the expression obtained, whether the stopping distance depends on the mass of the puck. What practical implication does this result have for road safety?

A cable car cabin with its passengers has a total mass of $800\ \text{kg}$. On the outbound journey, it ascends from the base station, at an altitude of $1200\ \text{m}$, to the top station, at $1750\ \text{m}$. It then partially descends to an intermediate station located at $1550\ \text{m}$ (take $g = 10\ \text{m/s}^2$), as shown in the figure.

1. a) Calculate the work done by weight during the ascent.
2. b) Calculate the work done by weight during the partial descent.
3. c) Calculate the total work done by weight over the entire journey and verify that it depends only on the net difference in altitude between the starting and ending points. What property of the gravitational force does this result highlight?

An elevator cabin with its passengers has a mass of $600\ \text{kg}$. It starts from rest and ascends $12\ \text{m}$ with a constant acceleration of $1,2\ \text{m/s}^2$ (take $g = 10\ \text{m/s}^2$).

1. a) Calculate the force exerted by the cable applying Newton's second law: $T - m \cdot g = m \cdot a$.
2. b) Calculate the work done by the cable and the work done by weight during these $12\ \text{m}$.
3. c) Find the net work and, using the work-energy theorem, the speed of the cabin after $12\ \text{m}$. Check the result using kinematics: $v^2 = 2 \cdot a \cdot h$.