First slide

Second law

Question

A balloon starting from rest ascends vertically with uniform acceleration to a height of 100 m in 10 s. The force on the bottom of the balloon by a mass of 50 kg is (Take, g = 10 ms^-2) [EAMCET 2013]

Moderate

A

100 N
B

300 N
C

600 N
D

400 N

Solution

Given, h = 100 m and t = 10 s
From second equation of motion, $h = \frac{1}{2} a t^{2}$
$\Rightarrow a = \frac{2 h}{t^{2}} = \frac{2 \times 100}{(10)^{2}} = 2 {ms}^{- 2}$
Now, $F = m (g + a) = 50 (10 + 2) = 600 N$

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