A liquid drop of 1.00 g falls from height of cliff 1.00 km. It hits the ground with a speed of 50 m s^-1. What is the work done by the unknown force?
A liquid drop of 1.00 g falls from height of cliff 1.00 km. It hits the ground with a speed of 50 m s-1. What is the work done by the unknown force? (Take g = 9.8 m/s2)
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Given: m = 1.0 g = 1.0 × 10-3 kg,
h = 1 km = 103 m, v = 50 ms-1
To find: Work done (Wf)
Formula: Wf = ∆ K.E – Wg
Calculation:
i. The change in kinetic energy of the drop
∆ K.E = (K.E.)final (K.E.)initial
∴ ∆ K.E. = \(\frac{1}{2}\)mv2 − 0
= \(\frac{1}{2}\) × 1.0 × 10-3 × (50)2
∴ ∆ K.E.= 1.25 J
ii. Work done by the gravitational force is,
Wg = mgh = 1.0 × 10-3 × 9.8 × 103 = 9.8 J
∴ Wg = 9.8J
From formula,
Wf = ∆K.E. – Wg = 1.25 – 9.8
Wf = -8.55 J
Work done by the unknown force is – 8.55 J.
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