Worked examples · Sheet 2

Speeds and stopping distances with the work-energy theorem

Theory used: Work-energy theorem Kinetic energy Work of the net force Energy dissipated by friction

Go to: Exercise 1 Exercise 2 Exercise 3 Exercise 4

Exercise 1

A car of mass $m = 1200 kg$ travels at $72 km/h$ on a horizontal road and brakes with a constant braking force of magnitude $F = 6000 N$ . Compute the braking distance.

L_{tot} = - F d, L_{tot} = K_{f} - K_{i} = 0 - \frac{1}{2} m v_{0}^{2}

- F d = - \frac{1}{2} m v_{0}^{2} ⟹ d = \frac{m v _{0}^{2}}{2 F}

v_{0} = \frac{72}{3.6} m/s = 20 m/s, d = \frac{1200 \cdot 2 0 ^{2}}{2 \cdot 6000} m = 40 m

The forces acting on the car are the weight, the normal force and the braking force. The weight and the normal force are vertical and the displacement is horizontal, so they do no work; the braking force is opposite to the displacement and does the work $- F d$ . By the work-energy theorem this work equals the change in kinetic energy, which goes from the initial value $\frac{1}{2} m v_{0}^{2} = 2.4 \cdot 1 0^{5} J$ to zero. The speed must be expressed in meters per second before substituting, by dividing the value in kilometers per hour by $3.6$ . The braking distance is therefore $40 m$ . As a dimensional check, $\frac{kg \cdot m ^{2} / s ^{2}}{N} = m$ .

Exercise 2

A block of mass $m = 2.0 kg$ , initially at rest on a horizontal plane, is pulled by a constant horizontal force $F = 15 N$ over a distance $d = 4.0 m$ . The coefficient of kinetic friction is $μ_{d} = 0.30$ . Compute the speed of the block at the end of the distance.

L_{tot} = (F - μ_{d} m g) d, L_{tot} = \frac{1}{2} m v^{2} - 0

v = \frac{2 ( F - μ _{d} m g ) d}{m}

μ_{d} m g L_{tot} v = 0.30 \cdot 2.0 \cdot 9.81 N \approx 5.89 N = (15 - 5.89) \cdot 4.0 J \approx 36.5 J = \frac{2 \cdot 36.5}{2.0} m/s \approx 6.0 m/s

The plane is horizontal and the force $F$ is horizontal, so the normal force balances the weight, $N = m g$ , and kinetic friction has magnitude $μ_{d} m g$ . The force $F$ is in the direction of the displacement and friction is opposite to it, while the weight and the normal force do no work. By the work of the net force, the total work is that of the net force $F - μ_{d} m g$ over the distance $d$ . The block starts from rest, therefore by the work-energy theorem the total work equals the final kinetic energy, from which the speed is obtained: $v \approx 6.0 m/s$ .

Exercise 3

A bullet of mass $m = 10 g$ strikes a wooden board $s = 5.0 cm$ thick at speed $v_{A} = 400 m/s$ and exits at speed $v_{B} = 150 m/s$ . Compute the magnitude of the average force that the wood exerts on the bullet, assuming straight-line motion and neglecting the weight during the crossing.

- \overset{ˉ}{F} s = \frac{1}{2} m v_{B}^{2} - \frac{1}{2} m v_{A}^{2} ⟹ \overset{ˉ}{F} = \frac{m ( v _{A}^{2} - v _{B}^{2} )}{2 s}

K_{B} - K_{A} \overset{ˉ}{F} = \frac{1}{2} \cdot 0.010 \cdot (15 0^{2} - 40 0^{2}) J = - 687.5 J = \frac{687.5}{0.050} N \approx 1.4 \cdot 1 0^{4} N

The force of the wood on the bullet is not constant, and how it varies is not known. We then define the average force $\overset{ˉ}{F}$ as the constant force that, over the same distance $s$ , would do the same work; since the force opposes the motion, the work is $- \overset{ˉ}{F} s$ . The weight is neglected, as required, so the only force that does work is that of the wood. By the work-energy theorem the work equals the change in kinetic energy, negative because the bullet slows down. Mass and thickness must be converted to kilograms and meters. The average force is about $1.4 \cdot 1 0^{4} N$ , about one hundred forty thousand times the weight of the bullet, which is $0.098 N$ .

Exercise 4

A block is launched with speed $v_{0} = 6.0 m/s$ up a plane inclined at $α = 2 0^{\circ}$ . The coefficient of kinetic friction is $μ_{d} = 0.15$ and the coefficient of static friction is $μ_{s} = 0.30$ . Compute the distance traveled along the plane before stopping and determine whether the block stays at rest or slides back.

L_{tot} = - m g sin α d - μ_{d} m g cos α d, L_{tot} = 0 - \frac{1}{2} m v_{0}^{2}

d = \frac{v _{0}^{2}}{2 g ( sin α + μ _{d} cos α )}

sin α + μ_{d} cos α d = 0.342 + 0.15 \cdot 0.940 \approx 0.483 = \frac{6. 0 ^{2}}{2 \cdot 9.81 \cdot 0.483} m \approx 3.8 m

tan α = tan 2 0^{\circ} \approx 0.364 > μ_{s} = 0.30

During the ascent the weight, the normal force and kinetic friction act. The block rises by the height $d sin α$ , so the weight does the work $- m g d sin α$ . The normal force is $m g cos α$ and does no work; friction, of magnitude $μ_{d} m g cos α$ , points down the incline, opposite to the displacement, and does negative work. At the highest point the speed is zero, therefore by the work-energy theorem the sum of the two works equals $- \frac{1}{2} m v_{0}^{2}$ . The mass cancels and we obtain $d \approx 3.8 m$ .

At rest at the highest point, the block stays in equilibrium only if static friction can balance the component of the weight along the plane, that is, if $m g sin α \leq μ_{s} m g cos α$ , which is equivalent to $tan α \leq μ_{s}$ . Since $tan 2 0^{\circ} \approx 0.364$ exceeds $μ_{s} = 0.30$ , the condition is not satisfied, so the block slides back.