Question #5fc94

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Question #5fc94

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Answer 1 · 2017-10-13T23:44:47

The frictional force is 9.8 N.

Explanation:

I will assume 1) this mass is on a horizontal surface and 2) the 70 N force is applied horizontally.

The formula for the frictional force, $F_{f}$ $F_f$ , is

$F_{f} = μ \cdot N$ $F_f = mu*N$

where $μ$ $mu$ is the coefficient of friction and N is Normal force. Normal force means the force pressing the object to the surface it is sliding on. In this case, it is the weight of the mass, so $N = m \cdot g$ $N = m*g$ .

$F_{f} = 0.02 \cdot 50 k g \cdot 9.8 \frac{m}{s^{2}} = 9.8 N$ $F_f = 0.02*50 kg*9.8 m/s^2 = 9.8 N$

The net force will be
$70 N - 9.8 N = 60.2 N$ $70 N - 9.8 N = 60.2 N$
The friction will not be large enough to prevent the mass from sliding, so the mass will accelerate.

I hope this helps,
Steve

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Question #5fc94

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