A box with an initial speed of #7 m/s# is moving up a ramp. The ramp has a kinetic friction coefficient of #1/3 # and an incline of #( pi )/3 #. How far along the ramp will the box go?

Answer 1

Olivia Askew

#l=2,41 " meters"#

#"total energy ":E_k=1/2*m*v^2" (at A)"#

#"energy changed heat due to friction :" W_f=mu* m*g*cos (pi/3)*l#

#"potential energy for the point of A :"E_p=m*g*h#

#"we can write the energy conservation equation as :"#

#E_k=E_p+W_f#

#1/2*m*v^2=m*g*h+mu*m*g*cos (pi/3)*l#

#"so ;"h=l.sin (pi/3)#

#1/2*cancel(m)*v^2=cancel(m)*g*l*sin(pi/3)+mu*cancel(m)*g*cos (pi/3)*l#

#1/2*v^2=g*l*0,87+1/3*g*0,5*l#

#v^2=g*l(2*0,87+(2*0,5)/3)#

#v=7 m/s#

#7^2=g*l(1,74+1/3)#

#49=g*l((5,22+1)/3)#

#49=g*l*2,07#

#g=9,81 m/s^2#

#49=9,81*l*2,07#

#49=20,31*l#

#l=2,41 " meters"#

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Answer 2

Adrian Ayala

To determine how far the box will go up the ramp, you can use the following equation:

$d = \frac{v_i^2}{2g} \left(1 - \frac{\mu_k}{\tan(\theta)}\right)$

where: $d$ is the distance along the ramp, $v_i$ is the initial speed (7 m/s), $g$ is the acceleration due to gravity (approximately 9.8 m/s²), $\mu_k$ is the kinetic friction coefficient (1/3), $\theta$ is the incline angle ( $\pi/3$ ).

Plug in the values and calculate the distance $d$ .

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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.