# How much work would it take to horizontally accelerate an object with a mass of #8 kg# to #3 m/s# on a surface with a kinetic friction coefficient of #1 #?

Concept 1 In the problem, it is important to note that there is no friction. In the formula,

Concept 2 To solve the problem, you must use two formulas:

Substitute your values.

Solve.

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To calculate the work done to horizontally accelerate an object to a certain velocity against friction, we can use the work-energy principle.

The work done is equal to the change in kinetic energy of the object:

[ \text{Work} = \Delta KE ]

[ \text{Work} = \frac{1}{2} m (v_f^2 - v_i^2) ]

Where:

- ( m ) is the mass of the object (8 kg),
- ( v_f ) is the final velocity (3 m/s),
- ( v_i ) is the initial velocity (assumed to be 0 m/s since we're starting from rest).

Plugging in the values:

[ \text{Work} = \frac{1}{2} \times 8 \times (3^2 - 0^2) ]

[ \text{Work} = \frac{1}{2} \times 8 \times (9 - 0) ]

[ \text{Work} = \frac{1}{2} \times 8 \times 9 ]

[ \text{Work} = 36 \text{ J} ]

So, it would take 36 Joules of work to horizontally accelerate the object to a velocity of 3 m/s against a surface with a kinetic friction coefficient of 1.

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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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