# A box with an initial speed of #6 m/s# is moving up a ramp. The ramp has a kinetic friction coefficient of #4/5 # and an incline of #( 5 pi )/12 #. How far along the ramp will the box go?

The distance is

Consequently, the object's net force is

Newton's Second Law states

A deceleration is indicated by the negative sign.

We utilize the equation of motion.

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To find how far along the ramp the box will go, you can use the work-energy principle. The work done by friction will be equal to the change in kinetic energy of the box.

The work done by friction is given by: [ W_{friction} = \mu_k \cdot m \cdot g \cdot d \cdot \cos(\theta) ] where:

- ( \mu_k ) is the coefficient of kinetic friction (4/5),
- ( m ) is the mass of the box (not given),
- ( g ) is the acceleration due to gravity (9.8 m/s²),
- ( d ) is the distance along the ramp,
- ( \theta ) is the angle of the incline (5π/12 radians).

The change in kinetic energy is given by: [ \Delta KE = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2 ] where:

- ( v_f ) is the final velocity (0 m/s, as the box stops at the top of the ramp),
- ( v_i ) is the initial velocity (6 m/s).

Setting the work done by friction equal to the change in kinetic energy and solving for ( d ), we get: [ \mu_k \cdot m \cdot g \cdot d \cdot \cos(\theta) = \frac{1}{2} m v_i^2 ] [ d = \frac{\frac{1}{2} m v_i^2}{\mu_k \cdot g \cdot \cos(\theta)} ]

We can cancel out the mass ( m ) from both sides of the equation, as it appears in both numerator and denominator.

[ d = \frac{\frac{1}{2} v_i^2}{\mu_k \cdot g \cdot \cos(\theta)} ]

Substituting the given values: [ d = \frac{\frac{1}{2} \times 6^2}{\frac{4}{5} \cdot 9.8 \cdot \cos\left(\frac{5\pi}{12}\right)} ]

[ d = \frac{\frac{1}{2} \times 36}{\frac{4}{5} \cdot 9.8 \cdot \cos\left(\frac{5\pi}{12}\right)} ]

[ d = \frac{18}{\frac{4}{5} \cdot 9.8 \cdot \cos\left(\frac{5\pi}{12}\right)} ]

[ d = \frac{18}{\frac{4}{5} \cdot 9.8 \cdot \cos\left(\frac{5\pi}{12}\right)} ]

[ d = \frac{18}{\frac{4}{5} \cdot 9.8 \cdot \cos\left(\frac{5\pi}{12}\right)} ]

[ d ≈ \frac{18}{7.84 \cdot \cos\left(\frac{5\pi}{12}\right)} ]

[ d ≈ \frac{18}{7.84 \cdot \cos\left(\frac{5\pi}{12}\right)} ]

[ d ≈ \frac{18}{7.84 \cdot 0.2588} ]

[ d ≈ \frac{18}{2.025} ]

[ d ≈ 8.89 ]

So, the box will travel approximately 8.89 meters along the ramp.

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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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