A $15$ $15$ $k g$ $kg$ block of steel is at rest on a smooth, horizontal, icy surface. What net force must applied to the block so that it accelerates at $0.6$ $0.6$ $m$ $m$ $/$ $/$ $s^{2}$ $s^2$ ?

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A $15$ $15$ $k g$ $kg$ block of steel is at rest on a smooth, horizontal, icy surface. What net force must applied to the block so that it accelerates at $0.6$ $0.6$ $m$ $m$ $/$ $/$ $s^{2}$ $s^2$ ?

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Answer 1 · 2016-03-06T21:49:52

$F_{n e t} = 9$ $F_{n et} =9$ $N$ $N$

Explanation:

The question asks for the required net force for a particular acceleration. The equation that relates the net force to the acceleration is Newton's 2nd Law,

$F_{n e t} = m a$ $F_{n et} =m a$ ,

where $F_{n e t}$ $F_{n et}$ is the net force normally in Newtons, $N$ $N$ ;

$m$ $m$ is the mass, in kilograms, $k g$ $kg$ ;

and $a$ $a$ is the acceleration in meters per second squared, $\frac{m}{s^{2}}$ $m/s^2$ .

We have $m = 15$ $m=15$ $k g$ $kg$ and $a = 0.6$ $a=0.6$ $\frac{m}{s^{2}}$ $m/s^2$ , so

$F_{n e t} = (15$ $F_{n et} =(15$ $k g) \cdot (0.6$ $kg )* (0.6$ $\frac{m}{s^{2}}$ $m/s^2$ $) = (15 \cdot 0.6) \cdot (k g \cdot \frac{m}{s^{2}})$ $)= (15*0.6) * (kg*m/s^2 )$

remember $1$ $1$ $N = k g \cdot \frac{m}{s^{2}}$ $N=kg*m/s^2$

$F_{n e t} = 9$ $F_{n et} =9$ $N$ $N$

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A $15$ $15$ $k g$ $kg$ block of steel is at rest on a smooth, horizontal, icy surface. What net force must applied to the block so that it accelerates at $0.6$ $0.6$ $m$ $m$ $/$ $/$ $s^{2}$ $s^2$ ?

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Explanation:

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A $15$ $15$ $k g$ $kg$ block of steel is at rest on a smooth, horizontal, icy surface. What net force must applied to the block so that it accelerates at $0.6$ $0.6$ $m$ $m$ $/$ $/$ $s^{2}$ $s^2$ ?