Question

astronauts in space cannot weigh themselves by standing on a bathroom scale. Instead, they determine their mass by oscillating on a large spring. Suppose an astronaut attaches one end of a large spring to her belt and the other end to a hook on the wall of the space capsule. A fellow astronaut then pulls her away from the wall and releases her. The spring's length as a

Answer 1

Answer:

The right answer is:

(a) 63.83 kg

(b) 0.725 m/s

Explanation:

The given query seems to be incomplete. Below is the attachment of the full question is attached.

The given values are:

T = 3 sec

k = 280 N/m

(a)

The mass of the string will be:

⇒ $T=2 \pi\sqrt{\frac{m}{k} }$

or,

⇒ $m=\frac{k T^2}{4 \pi^2}$

On substituting the values, we get

⇒     $=\frac{280\times (3)^2}{4 \pi^2}$

⇒     $=\frac{280\times 9}{4\times (3.14)^2}$

⇒     $=68.83 \ kg$

(b)

The speed of the string will be:

⇒  $\frac{1}{2}k(0.4)^2=\frac{1}{2}k(0.2)^2+\frac{1}{2}mv^2$

then,

⇒             $v=\sqrt{\frac{k((0.4)^2-(0.2)^2)}{m} }$

On substituting the values, we get

⇒                $=\sqrt{\frac{280\times ((0.4)^2-(0.2)^2)}{63.83} }$

⇒                $=\sqrt{\frac{280(0.16-0.04)}{63.83} }$

⇒                $=\sqrt{\frac{280\times 0.12}{63.83} }$

⇒                $=0.725 \ m/s$