An object with a mass of #5 # # kg# is on a surface with a kinetic friction coefficient of # 4 #. How much force is necessary to accelerate the object horizontally at # 6# #ms^-2#?

Answer 1

#F=F_"frict"+F_"accel"=196+30=226# #N#

The total force required will be given by #F=F_"frict"+F_"accel"# - the sum of the force required to overcome friction and the force required to accelerate the mass.

The frictional force is given by:

#F_"frict"=muF_"norm"# where the normal force #F_"norm"=mg#
So the friction force is #F_"frict"=mumg=4xx5xx9.8=196# #N#

To accelerate a mass the force given by Newton's Second Law is:

#F_"accel"=ma=5xx6=30# #N#

Then:

#F=F_"frict"+F_"accel"=196+30=226# #N#
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Answer 2

To determine the force necessary to accelerate the object horizontally at (6 , \text{m/s}^2), we need to consider both the force of kinetic friction and the force required for acceleration.

First, calculate the force of kinetic friction using the formula:

(F_{\text{friction}} = \mu_k \times N),

where (\mu_k) is the kinetic friction coefficient and (N) is the normal force.

Next, calculate the force required for acceleration using Newton's second law:

(F_{\text{net}} = m \times a),

where (m) is the mass of the object and (a) is the acceleration.

Finally, add the force of kinetic friction to the force required for acceleration to find the total force:

(F_{\text{total}} = F_{\text{friction}} + F_{\text{net}}).

Substitute the given values:

(F_{\text{friction}} = 4 \times 5 , \text{kg} \times 9.8 , \text{m/s}^2) (assuming standard gravity)
(F_{\text{net}} = 5 , \text{kg} \times 6 , \text{m/s}^2)

Then, calculate:

(F_{\text{friction}} = 4 \times 5 \times 9.8 = 196 , \text{N})
(F_{\text{net}} = 5 \times 6 = 30 , \text{N})

Finally, sum the forces:

(F_{\text{total}} = 196 + 30 = 226 , \text{N}).

Therefore, the force necessary to accelerate the object horizontally at (6 , \text{m/s}^2) is (226 , \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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