Work and Power
5 marks · NormalA small electric cart has a mass of 120 kg and moves at a steady speed of 5 m/s along a horizontal track for 40 m. The total frictional force opposing its motion is 90 N.
Calculate the work done against friction as the cart moves 40 m.
Calculate the kinetic energy of the cart as it moves.
State whether the work done against friction is equal to, less than, or greater than the kinetic energy of the cart. Explain your answer.
Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.
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(a) 3600 J
(b) 1500 J
(c) The work done against friction is greater than the kinetic energy. The cart loses more energy to friction than it has as kinetic energy.
Also accept: For part (c): "Greater than, because the work done against friction is the energy lost to the surroundings as heat/sound, while the kinetic energy is the energy of motion. For the same movement, work done against friction can be more/less/equal depending on forces, but in this case, their values can be compared quantitatively."
Common mistakes: Confusing force with work done; Mixing up kinetic energy and work done values
Show mark scheme and worked solution
Marking points
M1 (a) · 1 mark · Correctly identifies W = F × d
A1 (a) · 1 mark · Correct substitution and value for work done (3600 J) (follow-through)
M2 (b) · 1 mark · Uses KE = 1/2 mv^2 correctly
A2 (b) · 1 mark · Correct substitution and value for KE (1500 J) (follow-through)
B1 (c) · 1 mark · Correct comparative statement with explanation
Worked solution
a) Work done against friction = force × distance = 90 N × 40 m = 3600 J
b) KE = (1/2) × 120 × (5^2) = (1/2) × 120 × 25 = 60 × 25 = 1500 J
c) 3600 J (work done against friction) is greater than 1500 J (kinetic energy), so the cart loses more energy to friction than it has as KE.