4.0 kg |2068
The system shown above is released from rest. If friction is negligiths, the acceleration of the 4.0 kg block sliding on the table shown above is most nearly
0
B) 1.7 m/s2
С
3.3 m/s2
D) 5.0 m/s2
E
10.0 m/s2

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Ответ:

jay0626

02.04.2020

Physics

8.04924 m/s²

Explanation:

r = Distance the child is from the center = 4.65 m

$\alpha$ = Angular acceleration = 0.745 rad/s²

$\omega$ = Angular velocity = 1.25 rad/s

Velocity is given by

$v=r\omega\\\Rightarrow v=4.65\times 1.25\\\Rightarrow v=5.8125\ m/s$

Radial acceleration is given by

$a_r=\frac{v^2}{r}\\\Rightarrow a_r=\frac{5.8125^2}{4.65}\\\Rightarrow a_r=7.265625\ m/s^2$

Tangential acceleration is given by

$a_t=\alpha r\\\Rightarrow a_t=0.745\times 4.65\\\Rightarrow a_t=3.46425\ m/s^2$

The resultant acceleration is given by

$a=\sqrt{a_r^2+a_t^2}\\\Rightarrow a=\sqrt{7.265625^2+3.46425^2}\\\Rightarrow a=8.04924\ m/s^2$

The magnitude of the acceleration of the child is 8.04924 m/s²

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4.0 kg |2068 The system shown above is released from rest. If friction is negligiths, the acceleration of the 4.0 kg block sliding on the table shown above is most nearly 0 B) 1.7 m/s2 С 3.3 m/s2 D) 5.0 m/s2 E 10.0 m/s2