A 20.7 kg shopping trolley full of groceries, with frictionless wheels, starting from rest is pushed 13 cm in 0.05 s with a constant acceleration. How much work is done on it? Give your answer to two significant figures.

Question

BlackTom AI · Accepted Answer

Work done equals the change in kinetic energy, W = ΔK. With frictionless wheels and the trolley starting from rest, all the work done goes into the trolley's kinetic energy (initial KE = 0), so W = 1/2 m v^2.

Relevant formulae and concepts:
- Constant-acceleration kinematics: s = (1/2) a t^2 (starting from rest)
- Acceleration a = 2s / t^2
- Final speed v = a t
- Work-energy theorem: W = ΔK = (1/2) m v^2 (no energy lost to friction)

Step-by-step calculation (showing intermediate steps):
1. Convert units: s = 13 cm = 0.13 m; t = 0.05 s; m = 20.7 kg.
2. Find acceleration using s = 1/2 a t^2:
   a = 2s / t^2 = 2(0.13 m) / (0.05 s)^2 = 0.26 / 0.0025 = 104 m/s^2.
3. Find final speed: v = a t = 104 m/s^2 × 0.05 s = 5.2 m/s.
4. Compute kinetic energy (and thus work done):
   W = 1/2 m v^2 = 0.5 × 20.7 kg × (5.2 m/s)^2.
   v^2 = 5.2^2 = 27.04.
   W = 0.5 × 20.7 × 27.04 = 10.35 × 27.04 = 279.864 J.

Alternate direct check: W = F s = m a s = m (2s/t^2) s = 2 m s^2 / t^2 = 279.864 J (same result).

Why this is correct:
- The frictionless condition ensures no energy is lost; the kinematic relations correctly give the acceleration and final speed from the displacement and time; the work-energy theorem then yields the work as the kinetic energy gained.

Common mistakes to avoid:
- Forgetting to convert cm to m (would give a factor-100 error).
- Using incorrect kinematic formulae or mixing up v and a.
- Neglecting that initial KE = 0.

Rounded to two significant figures, W = 2.8 × 10^2 J, which matches the provided answer 2.8e2.

A 20.7 kg shopping trolley full of groceries, with frictionless wheels, starting from rest is pushed 13 cm in 0.05 s with a constant acceleration. How much work is done on it? Give your answer to two significant figures.数值题

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