Worked examples · Sheet 1

Work of constant forces

Theory used: Work of a constant force Sign of the work Work of the net force Work of the weight

Exercise 1

A crate is dragged for $d = 8.0 m$ across a horizontal floor by a rope that forms an angle $θ = 2 5^{\circ}$ with the horizontal. The tension in the rope is constant and equal to $T = 120 N$ . Compute the work done by the tension, by the weight and by the normal force.

L_{T} = T \cdot s = T d cos θ, L_{P} = 0, L_{N} = 0

L_{T} = 120 \cdot 8.0 \cdot cos 2 5^{\circ} J = 960 \cdot 0.906 J \approx 870 J

The tension is constant and the displacement is straight, so the definition of the work of a constant force applies: the dot product of force and displacement, equal to the product of the magnitudes times the cosine of the angle between them. The angle between the rope and the horizontal displacement is $2 5^{\circ}$ , and its cosine is positive, therefore the tension does positive work. The weight is vertical and the normal force is vertical, while the displacement is horizontal; since both forces form a right angle with the displacement, their work is zero. The result $870 J$ has two significant figures, like the data $8.0 m$ and $2 5^{\circ}$ .

Exercise 2

A sled of mass $m = 15 kg$ is pulled for $d = 12 m$ along a horizontal snowy track by a constant force $F = 80 N$ inclined upward at $θ = 3 0^{\circ}$ to the horizontal. The coefficient of kinetic friction between sled and snow is $μ_{d} = 0.10$ . Compute the work of each force and the total work.

Forces on the sled:

F

inclined at

θ = 3 0^{\circ}

, the normal force

N

, the weight

P

and the friction

f

opposite to the displacement, which is to the right

{N + F sin θ - m g = 0 f = μ_{d} N ⟹ f = μ_{d} (m g - F sin θ)

L_{F} = F d cos θ, L_{f} = - μ_{d} (m g - F sin θ) d, L_{P} = L_{N} = 0

N L_{F} L_{f} L_{tot} = (15 \cdot 9.81 - 80 \cdot 0.50) N = (147.15 - 40) N = 107.15 N = 80 \cdot 12 \cdot cos 3 0^{\circ} J \approx 831.4 J = - 0.10 \cdot 107.15 \cdot 12 J \approx - 128.6 J = L_{F} + L_{f} + L_{P} + L_{N} \approx 702.8 J \approx 7.0 \cdot 1 0^{2} J

To compute the friction we need the normal force. The sled does not move vertically, so the sum of the vertical components of the forces is zero; the force $F$ has a vertical component $F sin θ$ pointing upward, therefore the normal force is smaller than the weight, $N = m g - F sin θ$ . Kinetic friction has magnitude $μ_{d} N$ and is opposite to the displacement, so it forms an angle of $18 0^{\circ}$ with it and does negative work. The weight and the normal force are perpendicular to the displacement and do zero work. By the work of the net force, the total work is the sum of the four works; with two significant figures, like the data, it equals $7.0 \cdot 1 0^{2} J$ .

Exercise 3

A block of mass $m = 4.0 kg$ is pushed for $d = 3.0 m$ up a plane inclined at $α = 3 0^{\circ}$ by a force $F = 40 N$ parallel to the plane. The coefficient of kinetic friction is $μ_{d} = 0.25$ . Compute the work of the force $F$ , of the weight, of the normal force, of friction, and the total work.

The block moves up the plane:

F

is parallel to the plane pointing up, the friction

f

is parallel to the plane pointing down

L_{F} = F d, L_{P} = - m g d sin α, L_{N} = 0, L_{f} = - μ_{d} m g cos α d

L_{F} L_{P} L_{f} L_{tot} = 40 \cdot 3.0 J = 120 J = - 4.0 \cdot 9.81 \cdot 3.0 \cdot 0.50 J \approx - 58.9 J = - 0.25 \cdot 4.0 \cdot 9.81 \cdot 0.866 \cdot 3.0 J \approx - 25.5 J = (120 - 58.9 - 25.5) J \approx 35.7 J

The force $F$ is parallel to the displacement and in the same direction, so its work is the product of the magnitudes. The weight forms with the displacement an angle of $9 0^{\circ} + 3 0^{\circ} = 12 0^{\circ}$ , and $cos 12 0^{\circ} = - sin 3 0^{\circ}$ ; equivalently, the block rises by the height $d sin α = 1.5 m$ , therefore by the work of the weight we have $L_{P} = - m g d sin α$ . The normal force is perpendicular to the plane and does no work. In the direction perpendicular to the plane there is no motion and $F$ has no component in that direction, so $N = m g cos α$ and friction has magnitude $μ_{d} m g cos α$ ; it is opposite to the displacement and does negative work. The total work, the sum of the four contributions, is positive, so the block speeds up as it moves up.