A $9.7 kg$ $9.7"kg"$ mass is positioned on top of a vertical spring. What is the upward force exerted on the mass by the spring?

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Answer 1 · 2016-11-20T21:30:41

$F_{s p} = 85 N$ $F_(sp)=85N$

Use Newton's second law, which (in summary) states that $F = m a$ $F=ma$ . Assuming no friction and that the mass is given in kilograms,

$F_{s p} = 9.7 k g \cdot 8.8 \frac{m}{s^{2}}$ $F_(sp)=9.7kg*8.8m/s^2$

$F_{s p} = 85 N$ $F_(sp)=85N$

Answer 2 · 2016-11-27T02:27:55

Net downward force on the 9.7 kg mass $F_{n} = 9.7 \times 8.8 N$ $F_n= 9.7xx8.8N$

Gravitational force on the 9.7 kg mass $F_{g} = 9.7 \times 9.8 N$ $F_g= 9.7xx9.8N$

If the upward force exerted by spring be $F_{s}$ $F_s$ then

$F_{n} = F_{g} - F_{s}$ $F_n=F_g-F_s$

$\Rightarrow F_{s} = F_{g} - F_{n} = 9.7 \times 9.8 - 9.7 \times 8.8 = 9.7 N$ $=>F_s=F_g-F_n=9.7xx9.8-9.7xx8.8=9.7N$

A 9.7"kg"9.7kg9.7"kg" mass is positioned on top of a vertical spring. What is the upward force exerted on the mass by the spring?