A 47.8 kg shopping trolley full of groceries, with frictionless wheels, starting from rest is pushed 18 cm in 0.19 s with a constant acceleration. How much work is done on it? Give your answer to two significant figures.

Question

BlackTom AI · Accepted Answer

A moving object under constant acceleration from rest allows us to connect displacement, time, and acceleration. Start with s = (1/2) a t^2 because the initial speed is zero. Solve for a: a = 2s / t^2 = 2 × 0.18 m / (0.19 s)^2. Numerically, (0.19)^2 = 0.0361, so a ≈ 0.36 / 0.0361 ≈ 9.97 m/s^2. Next, use v^2 = u^2 + 2 a s with u = 0 to find the final speed: v^2 = 0 + 2 × 9.97 × 0.18 ≈ 3.592 (m/s)^2, giving v ≈ 1.895 m/s. The work done on the trolley equals the gain in kinetic energy, W = ΔK = (1/2) m v^2. Substituting m = 47.8 kg and v^2 ≈ 3.592 gives W ≈ 0.5 × 47.8 × 3.592 ≈ 23.9 × 3.592 ≈ 85.9 J. When expressed with two significant figures, this is about 8.6 × 10^1 J, i.e., roughly 86 J. If you prefer a more qualitative check, the work should match the kinetic energy gained, and the numbers above are consistent with the given short displacement and rapid acceleration.

A 47.8 kg shopping trolley full of groceries, with frictionless wheels, starting from rest is pushed 18 cm in 0.19 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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