Work, energy and power: practice worksheet

1. Question 1warm-up · calculator

   A crate moves horizontally along a level floor. A rope pulls the crate with a constant force.
   Data:
   Pulling force: $36.0\ \mathrm{N}$
   Angle of rope above the horizontal: $35.0^\circ$
   Horizontal displacement: $4.50\ \mathrm{m}$

   (a)Calculate the work done by the pulling force.[2]

   Markscheme

   $W=36.0\times4.50\cos35.0^\circ$	1
   $W=133\ \mathrm{J}$	1

   Accept equivalent forms.

   Answer: $W=36.0\times4.50\cos35.0^\circ=133\ \mathrm{J}$

   (b)State the work done by the normal reaction on the crate.[1]

   Markscheme

   The normal reaction does zero work.

   1

   The normal reaction is perpendicular to the displacement.

   Answer: $0\ \mathrm{J}$

2. Question 2warm-up · calculator

   A laboratory slider moves along a horizontal guide. Its mass remains constant.
   Data:
   Mass: $0.800\ \mathrm{kg}$
   Initial momentum magnitude: $12.0\ \mathrm{kg\,m\,s^{-1}}$
   Net work subsequently done on the slider: $+70.0\ \mathrm{J}$

   (a)Determine the initial kinetic energy of the slider.[2]

   Markscheme

   $E_k=\dfrac{12.0^2}{2(0.800)}$	1
   $E_k=90.0\ \mathrm{J}$	1

   Answer: $E_k=\frac{12.0^2}{2(0.800)}=90.0\ \mathrm{J}$

   (b)Hence determine the final speed of the slider.[2]

   Markscheme

   $\dfrac12(0.800)v^2=90.0+70.0$	1
   $v=20.0\ \mathrm{m\,s^{-1}}$	1

   ECF from part (a).

   Answer: $v=\sqrt{\frac{2(90.0+70.0)}{0.800}}=20.0\ \mathrm{m\,s^{-1}}$

3. Question 3exam-style · calculator

   A trolley of mass $2.40\ \mathrm{kg}$ starts from rest at displacement $s=0$. The resultant force is parallel to its displacement. A force–displacement graph consists of straight lines joining the following points in order:
   $$\begin{array}{c|rrrr}s\,/\,\mathrm{m}&0&2.00&5.00&7.00\\F\,/\,\mathrm{N}&20.0&60.0&60.0&20.0\end{array}$$

   (a)Determine the net work done on the trolley between $s=0$ and $s=7.00\ \mathrm{m}$.[2]

   Markscheme

   $W=\dfrac{20.0+60.0}{2}(2.00)+60.0(3.00)+\dfrac{60.0+20.0}{2}(2.00)$	1
   $W=340\ \mathrm{J}$	1

   Accept any equivalent decomposition of the area under the graph.

   Answer: $W=\frac{20.0+60.0}{2}(2.00)+60.0(3.00)+\frac{60.0+20.0}{2}(2.00)=340\ \mathrm{J}$

   (b)Hence calculate the speed of the trolley at $s=7.00\ \mathrm{m}$.[2]

   Markscheme

   $\dfrac12(2.40)v^2=340$	1
   $v=16.8\ \mathrm{m\,s^{-1}}$	1

   ECF from part (a).

   Answer: $v=\sqrt{\frac{2(340)}{2.40}}=16.8\ \mathrm{m\,s^{-1}}$

   (c)Determine the instantaneous power transferred to the trolley by the resultant force at $s=7.00\ \mathrm{m}$.[2]

   Markscheme

   $P=20.0v$, using the speed at $s=7.00\ \mathrm{m}$.	1
   $P=337\ \mathrm{W}$	1

   ECF from part (b). Accept $336\ \mathrm{W}$ when the rounded speed $16.8\ \mathrm{m\,s^{-1}}$ is used.

   Answer: $P=20.0\sqrt{\frac{2(340)}{2.40}}=337\ \mathrm{W}$

4. Question 4exam-style · calculator

   A trolley is released from rest at point A on a track. The track is smooth from A to B and rough from B to C. Heights are measured above the same horizontal reference level. Air resistance is negligible.
   Data:
   Trolley mass: $0.650\ \mathrm{kg}$
   Height of A: $3.40\ \mathrm{m}$
   Height of B: $1.20\ \mathrm{m}$
   Height of C: $0.400\ \mathrm{m}$
   Speed at C: $6.00\ \mathrm{m\,s^{-1}}$
   Take $g=9.81\ \mathrm{m\,s^{-2}}$.

   (a)Show that the speed of the trolley at B is about $6.6\ \mathrm{m\,s^{-1}}$.[2]

   Markscheme

   $\dfrac12mv_B^2+mg(1.20)=mg(3.40)$	1
   $v_B=\sqrt{2(9.81)(3.40-1.20)}=6.57\ \mathrm{m\,s^{-1}}\approx6.6\ \mathrm{m\,s^{-1}}$	1

   A numerical substitution leading to the stated result is required.

   Answer: $v_B\approx6.6\ \mathrm{m\,s^{-1}}$

   (b)Determine the work done against friction between B and C.[3]

   Markscheme

   $E_B=\dfrac12mv_B^2+mg(1.20)=21.6801\ \mathrm{J}$	1
   $E_C=\dfrac12(0.650)(6.00)^2+(0.650)(9.81)(0.400)=14.2506\ \mathrm{J}$	1
   $W_{\mathrm{against\ friction}}=E_B-E_C=7.43\ \mathrm{J}$	1

   Accept determining $E_B$ directly from the gravitational potential energy at A. ECF from part (a); using $6.60\ \mathrm{m\,s^{-1}}$ gives $7.56\ \mathrm{J}$. The work done by friction is negative, but the requested work done against friction is positive.

   Answer: $W_{\mathrm{against\ friction}}=7.43\ \mathrm{J}$

   (c)Explain how energy is conserved between B and C when the trolley, track and Earth are treated as a closed system.[2]

   Markscheme

   Mechanical energy is transformed into thermal energy in the trolley and track. OWTTE	1
   The thermal energy increase equals the mechanical energy decrease, so the total energy of the closed system is unchanged. OWTTE	1

   Answer: The decrease in mechanical energy becomes thermal energy in the trolley and track, so the total energy of the closed system remains constant.

5. Question 5exam-style · calculator

   An electric motor raises a loaded platform vertically at constant speed. A constant resistive force acts downwards. The useful output is defined as the increase in gravitational potential energy of the platform and load.
   Data:
   Combined mass of platform and load: $48.0\ \mathrm{kg}$
   Upward speed: $0.320\ \mathrm{m\,s^{-1}}$
   Resistive force: $25.0\ \mathrm{N}$
   Electrical input power: $0.240\ \mathrm{kW}$
   Take $g=9.81\ \mathrm{m\,s^{-2}}$.

   (a)Determine the upward force exerted by the motor on the platform.[2]

   Markscheme

   $F=mg+25.0=48.0(9.81)+25.0$	1
   $F=496\ \mathrm{N}$	1

   The resultant force is zero because the speed is constant.

   Answer: $F=48.0(9.81)+25.0=496\ \mathrm{N}$

   (b)Calculate the mechanical power delivered by the motor to the platform, in W.[2]

   Markscheme

   $P=F(0.320)$	1
   $P=159\ \mathrm{W}$	1

   ECF from part (a).

   Answer: $P=495.88(0.320)=159\ \mathrm{W}$

   (c)Determine the efficiency of the lifting system.[2]

   Markscheme

   $\eta=\dfrac{mgv}{P_{\mathrm{input}}}=\dfrac{48.0(9.81)(0.320)}{240}$	1
   $\eta=0.628$ or $62.8\%$.	1

   Accept useful power calculated as the answer to part (b) minus $25.0(0.320)\ \mathrm{W}$, with ECF from part (b).

   Answer: $\eta=\frac{48.0(9.81)(0.320)}{240}=0.628=62.8\%$

   (d)A Sankey diagram represents the energy transfers during $20.0\ \mathrm{s}$ of lifting at this speed. It has one electrical input branch, one useful output branch and one branch for all wasted energy. Determine the energy labels, in kJ, for the input branch and the wasted-energy branch.[2]

   Markscheme

   $E_{\mathrm{input}}=0.240(20.0)=4.80\ \mathrm{kJ}$	1
   $E_{\mathrm{wasted}}=4.80-\dfrac{48.0(9.81)(0.320)(20.0)}{1000}=1.79\ \mathrm{kJ}$	1

   ECF from part (c) for wasted energy calculated using $E_{\mathrm{input}}(1-\eta)$. Wasted energy includes dissipation within the motor and work against the resistive force.

   Answer: Input: $4.80\ \mathrm{kJ}$; wasted energy: $1.79\ \mathrm{kJ}$.

6. Question 6challenge · calculator

   A testing sledge is launched along a horizontal guide by a compressed spring. The sledge starts from rest and remains in contact with the spring until the spring reaches its natural length. It then separates from the spring, travels a further $1.50\ \mathrm{m}$ horizontally and climbs a straight ramp to point T.
   A constant resistive force opposes the motion over the entire path, including during contact with the spring. Air resistance is negligible. The spring has negligible mass. Its force has magnitude $F=kx$, where $x$ is its compression.
   Data:
   Sledge mass: $1.20\ \mathrm{kg}$
   Spring constant: $640\ \mathrm{N\,m^{-1}}$
   Initial spring compression: $0.250\ \mathrm{m}$
   Resistive force: $2.00\ \mathrm{N}$
   Ramp length to T: $1.60\ \mathrm{m}$
   Vertical height of T above the horizontal guide: $0.800\ \mathrm{m}$
   Take $g=9.81\ \mathrm{m\,s^{-2}}$.

   (a)Show that the speed of the sledge when it separates from the spring is about $5.7\ \mathrm{m\,s^{-1}}$.[2]

   Markscheme

   $\dfrac12(640)(0.250)^2=\dfrac12(1.20)v^2+2.00(0.250)$	1
   $v=\sqrt{\dfrac{2(19.5)}{1.20}}=5.70\ \mathrm{m\,s^{-1}}\approx5.7\ \mathrm{m\,s^{-1}}$	1

   A numerical substitution leading to the stated result is required.

   Answer: $v\approx5.7\ \mathrm{m\,s^{-1}}$

   (b)Determine the speed of the sledge when the spring compression has decreased to $0.125\ \mathrm{m}$.[2]

   Markscheme

   $\dfrac12(1.20)v^2=\dfrac12(640)(0.250^2-0.125^2)-2.00(0.125)$	1
   $v=4.96\ \mathrm{m\,s^{-1}}$	1

   The distance travelled is the decrease in compression.

   Answer: $v=\sqrt{\frac{2\left[\frac12(640)(0.250^2-0.125^2)-2.00(0.125)\right]}{1.20}}=4.96\ \mathrm{m\,s^{-1}}$

   (c)Hence calculate the instantaneous power delivered by the spring to the sledge at a compression of $0.125\ \mathrm{m}$.[2]

   Markscheme

   $P=Fv=(640)(0.125)v$	1
   $P=397\ \mathrm{W}$	1

   ECF from part (b). This is the power delivered by the spring.

   Answer: $P=(640)(0.125)(4.958\ldots)=397\ \mathrm{W}$

   (d)Determine the acceleration of the sledge at a spring compression of $0.125\ \mathrm{m}$.[2]

   Markscheme

   $(1.20)a=640(0.125)-2.00$	1
   $a=65.0\ \mathrm{m\,s^{-2}}$, in the direction of motion.	1

   This part links energy and power to Newton's second law.

   Answer: $a=\frac{640(0.125)-2.00}{1.20}=65.0\ \mathrm{m\,s^{-2}}$, forwards.

   (e)Calculate the speed of the sledge at T.[2]

   Markscheme

   $\dfrac12(1.20)v_T^2=\dfrac12(640)(0.250)^2-2.00(0.250+1.50+1.60)-1.20(9.81)(0.800)$	1
   $v_T=2.54\ \mathrm{m\,s^{-1}}$	1

   Accept an energy equation starting at the separation point, with ECF from part (a). Using $5.70\ \mathrm{m\,s^{-1}}$ gives $2.54\ \mathrm{m\,s^{-1}}$.

   Answer: $v_T=\sqrt{\frac{2\left[20.0-2.00(0.250+1.50+1.60)-1.20(9.81)(0.800)\right]}{1.20}}=2.54\ \mathrm{m\,s^{-1}}$

   (f)The initial spring compression is now adjusted. The spring's natural-length position, the horizontal distance after separation and the ramp remain fixed. Determine the minimum initial compression required for the sledge to reach T.[2]

   Markscheme

   $\dfrac12(640)x^2=1.20(9.81)(0.800)+2.00(x+1.50+1.60)$, with zero kinetic energy at T.	1
   Solving $320x^2-2.00x-15.6176=0$ gives the positive root $x=0.224\ \mathrm{m}$.	1

   Accept a numerical solution. The resistive work during spring contact changes when the initial compression changes.

   Answer: $x=\frac{2.00+\sqrt{2.00^2+4(320)(15.6176)}}{640}=0.224\ \mathrm{m}$

Original questions and notes written by Pietro Meloni, PhD, from the IB syllabus. Not endorsed by or affiliated with the International Baccalaureate Organization. Found a mistake? Please tell me privately at pietro@pietromeloni.com. More questions on this topic: A. Space, time and motion practice.