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 $32 \frac{m}{s^{2}}$ $32 m/s^2$ ?

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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 $32 \frac{m}{s^{2}}$ $32 m/s^2$ ?

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Answer 1 · 2015-12-25T19:28:12

Let us use the $2^{nd}$ $2^"nd"$ Newton law in order to get this.

Explanation:

According to Newton $2^{nd}$ $2^"nd"$ law, the total force made on a body (i.e. the sum of all forces) is proportional to its acceleration, in this way:

$F_{Total} = \sum F = m a$ $F_"Total"= sum F = m a$

In our question, there are two forces:
- Our own force, which we are going to call $F_{ours}$ $F_"ours"$ .
- The friction force, which we are going to call $F_{fr}$ $F_"fr"$ .

Both forces must be summed up, although with opposite sign, because friction always acts against movement.

We know that friction force can be obtained by:

$F_{fr} = μ \cdot N$ $F_"fr" = mu cdot N$

with $μ$ $mu$ being the kinetic friction coefficient, and $N$ $N$ being the normal force (equals to the weigth of the body in our problem).
So:

$F_{fr} = μ \cdot N = μ \cdot p = μ \cdot (m g) = 4 \cdot (5 kg \cdot 9.8 {m/s}^{2}) = 196 N$ $F_"fr" = mu cdot N = mu cdot p = mu cdot (m g) = 4 cdot (5 "kg" cdot 9.8 "m/s"^2) = 196 "N"$

Now that we know this, we can rewrite $2^{nd}$ $2^"nd"$ Newton law as:

$\sum F = F_{ours} - F_{fr} = m a \to$ $sum F = F_"ours" - F_"fr" = m a rightarrow$
$\to F_{ours} = m a + F_{fr} = 5 kg \cdot 32 {m/s}^{2} + 196 N = 356 N$ $rightarrow F_"ours" = ma + F_"fr" = 5 "kg" cdot 32 "m/s"^2 + 196 "N" = 356 "N"$

So, we must make a force of 356 N in order to move our object with an acceleration of $32 {m/s}^{2}$ $32 "m/s"^2$ .

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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 $32 \frac{m}{s^{2}}$ $32 m/s^2$ ?

1 Answer

Explanation:

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An object with a mass of 5kg5 kg is on a surface with a kinetic friction coefficient of 4 4 . How much force is necessary to accelerate the object horizontally at 32ms2 32 m/s^2?

1 Answer

Explanation:

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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 $32 \frac{m}{s^{2}}$ $32 m/s^2$ ?