An object with a mass of $5 k g$ $5 kg$ is on a surface with a kinetic friction coefficient of $3$ $3$ . How much force is necessary to accelerate the object horizontally at $6 \frac{m}{s^{2}}$ $6 m/s^2$ ?

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An object with a mass of $5 k g$ $5 kg$ is on a surface with a kinetic friction coefficient of $3$ $3$ . How much force is necessary to accelerate the object horizontally at $6 \frac{m}{s^{2}}$ $6 m/s^2$ ?

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Answer 1 · 2018-06-18T05:36:40

177 N

Explanation:

friction force is
$F_{f} = μ \times m \times g = 3 \times 5 k g \times 9, 8 \frac{m}{s^{2}} = 147 N$ $F_f= mu xx m xx g= 3 xx 5 kg xx 9,8 m/s^2= 147 N$
by the second dynamic principle
$F - F - f = m \times a$ $F - F-f= m xx a$
F= 147 N + 5 kg xx 6 m/s^2= 177 N#

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An object with a mass of $5 k g$ $5 kg$ is on a surface with a kinetic friction coefficient of $3$ $3$ . How much force is necessary to accelerate the object horizontally at $6 \frac{m}{s^{2}}$ $6 m/s^2$ ?

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An object with a mass of $5 k g$ $5 kg$ is on a surface with a kinetic friction coefficient of $3$ $3$ . How much force is necessary to accelerate the object horizontally at $6 \frac{m}{s^{2}}$ $6 m/s^2$ ?