Question

(8c8p49) A 115g Frisbee is thrown from a point 1.00 m above the ground with a speed of 12.00 m/s. When it has reached a height of 2.00 m, its speed is 10.9674 m/s. How much work was done on the Frisbee by its weight

Answer 1

Answer:

Explanation:

mass of frisbee, m = 115 g = 0.115 kg

initial height, h = 1 m

final height, h' = 2 m

Work done on frisbee by its weight

W = - mg(h' - h)

W = - 0.115 x 9.8 x ( 2 - 1)

W = - 1.127 J

Answer 2

Answer:

The work done on the Frisbee is 1.36 J.

Explanation:

Given that,

Mass of Frisbee, m = 115 g = 0.115 kg

Initial speed of Frisbee, u = 12 m/s at a point 1 m above the ground

Final speed of Frisbee , v = 10.9674 m/s when it has reached a height of 2.00 m. Let W is the work done on the Frisbee by its weight. According to work energy theorem, the work done is equal to the change in its kinetic energy. So,

$W=\dfrac{1}{2}m(v^2-u^2)\\\\W=\dfrac{1}{2}\times0.115\times\left((10.9674)^{2}-(12)^{2})\right)\\\\W=-1.36\ J$

So, the work done on the Frisbee is 1.36 J. Hence, this is the required solution.