An object with a mass of $5 k g$ $5 kg$ is on a surface with a kinetic friction coefficient of $4$ $4$ . How much force is necessary to accelerate the object horizontally at $9 \frac{m}{s^{2}}$ $9 m/s^2$ ?

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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 $4$ $4$ . How much force is necessary to accelerate the object horizontally at $9 \frac{m}{s^{2}}$ $9 m/s^2$ ?

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Answer 1 · 2016-03-29T01:32:02

$F_{x} = 5 (9 - .4 \cdot 10) \approx 25 N$ $F_x=5(9-.4*10) ~~25N$

Explanation:

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Draw a free body diagram, FBD:
x-axis:
$\sum F_{x} = m a; F_{x} - F_{μ_{k}} = m a$ $sumF_x=ma; F_x-F_(mu_k) = ma$ Newton Law ===>(1)
Summation of all Forces $\Rightarrow m a$ $=> ma$

y-axis:
$\sum F_{y} = m a; F_{w} - N = 0; F_{w} = m g = N$ $sumF_y=ma; F_w-N = 0; F_w=mg=N$ Newton Law
$F_{μ_{k}} = μ_{k} \cdot N = μ_{k} \cdot m g$ $F_(mu_k)=mu_k *N= mu_k *mg$ ===>(2)

Insert (2) into (1):

$F_{x} - μ_{k} \cdot m g = m a$ $F_x - mu_k *mg = ma$ Solve for $F_{x}$ $F_x$
$F_{x} = m a - μ_{k} \cdot m g = m (a - μ_{k} g)$ $F_x = ma-mu_k *mg = m(a-mu_kg)$

Now $m = 5 k g; μ_{k} = .4; a = 9 \frac{m}{s^{2}}$ $m=5kg; mu_k = .4; a=9m/s^2$
$F_{x} = 5 (9 - .4 \cdot 10) \approx 25 N$ $F_x=5(9-.4*10) ~~25N$

A word of caution, a kinetic friction, $μ_{k} = 4$ $mu_k = 4$ is very large, I suspect you meant to say $.4$ $.4$ which reasonable. I assumed .4. If you want $μ_{k} = 4$ $mu_k = 4$ just correct for it.

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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 $4$ $4$ . How much force is necessary to accelerate the object horizontally at $9 \frac{m}{s^{2}}$ $9 m/s^2$ ?

1 Answer

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An object with a mass of 5 kg5kg5 kg is on a surface with a kinetic friction coefficient of 4 4 4 . How much force is necessary to accelerate the object horizontally at 9 m/s^29ms2 9 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 $4$ $4$ . How much force is necessary to accelerate the object horizontally at $9 \frac{m}{s^{2}}$ $9 m/s^2$ ?