Question

A 9.0 kg pink grapefruit is accelerated horizontally from rest to a velocity of 10.0 m/s in 5.00 s by a horizontal force. How much work is done on the pink grapefruit assuming no friction?

Answer 1

The work done on the pink grapefruit is 450 J.

The given parameters;

• mass of the pink grapefruit, m = 9 kg
• final velocity of the fruit, v = 10 m/s
• time of motion, t = 5 s

The acceleration of the grapefruit is calculated as follows;

$a = \frac{v}{t} \\\\a = \frac{10}{5} = 2 \ m/s^2$

The distance traveled by the grapefruit is calculated as follows;

$v^2 = u^2 + 2as\\\\v^2 = 0 + 2as\\\\v^2 = 2as\\\\s = \frac{v^2}{2a} \\\\s = \frac{(10)^2}{2(2)} \\\\s = 25 \ m$

The work done on the pink grapefruit is calculated as follows;

W = F x s

W = ma x s

W = (9 x 2) x 25

W = 450 J

Thus, the work done on the pink grapefruit is 450 J.

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