A liquid drop of 1.00 g falls from height of cliff 1.00 km. It hits th

1.	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)
Answer» Given: m = 1.0 g = 1.0 × 10^-3 kg, h = 1 km = 10³ m, v = 50 ms^-1 To find: Work done (W_f) Formula: W_f= ∆ K.E – W_g 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, W_g = mgh = 1.0 × 10^-3 × 9.8 × 10³ = 9.8 J ∴ W_g = 9.8J From formula, W_f= ∆K.E. – Wg = 1.25 – 9.8 W_f= -8.55 J Work done by the unknown force is – 8.55 J.

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)

Answer»

Given: m = 1.0 g = 1.0 × 10^-3 kg,

h = 1 km = 10³ m, v = 50 ms^-1

To find: Work done (W_f)

Formula: W_f= ∆ K.E – W_g

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,

W_g = mgh = 1.0 × 10^-3 × 9.8 × 10³ = 9.8 J

∴ W_g = 9.8J

From formula,

W_f= ∆K.E. – Wg = 1.25 – 9.8

W_f= -8.55 J

Work done by the unknown force is – 8.55 J.