An object with a mass of #6 kg# is on a surface with a kinetic friction coefficient of # 8 #. How much force is necessary to accelerate the object horizontally at # 7 m/s^2#?

Answer 1

Consider,

where #mu_k = 8.0# (which is like super glue!).

Recall Newton's second law,

#SigmaF_y = F_N - mg = 0#
#=>F_N = mg#

#therefore F_N = 6kg * (9.8m)/s^2 = 58.8N#

Hence,

#SigmaF_x = F_a - F_"fr" = ma#

#=> F_a - mu_k*F_N = ma#

#=> F_a = m(a+mu_k*F_N)#

#therefore F_a = 6kg * ((7m)/s^2 + 8.0*58.8N) approx 3kN#

That's some serious work we need to do!

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Answer 2

(F_{\text{net}} = m \cdot (a + g \cdot \mu_k))
(F_{\text{net}} = 6 , \text{kg} \cdot (7 , \text{m/s}^2 + 9.8 , \text{m/s}^2 \cdot 0.8))
(F_{\text{net}} = 6 , \text{kg} \cdot 15.24 , \text{m/s}^2)
(F_{\text{net}} = 91.44 , \text{N})

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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