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Car Accelerating Down a Hill & Continuing a Distance w/ Friction & Conservation of Energy

  1. Nov 16, 2014 #1
    1. The problem statement, all variables and given/known data
    A 900-kg car initially at rest rolls 50m down a hill inclined at an angle of 5.0^o. A 400-N effective friction force opposes its motion. (a) How fast is the car moving at the bottom? (b) What distance will it travel on a similar horizontal surface at the bottom of the hill? (c) Will the distance decrease or increase if the car’s mass is 1800-kg?

    [itex]m = 900 kg[/itex]
    [itex]\vec s_i = 50 m[/itex]
    [itex]\vec s_f = 0 m[/itex]
    [itex]\Delta \vec s = \vec s_f - \vec s_i = -50 m[/itex]
    [itex]\theta = 5.0^o[/itex]
    [itex]\vec F_f = 400 N[/itex]
    [itex]g = 9.8 m/s^2[/itex]
    [itex]KE_i = 0 J[/itex]
    [itex]PE_f = 0 J[/itex]

    2. Relevant equations
    My car is going down the hill and to the left. I choose up as positive, down as negative, right as positive and left as negative.

    (a) [itex]\sum \vec F_x = F_{net1} = F_f - F = F_f - mg sin \theta[/itex]
    [itex]W_{net} = {\vec F}_{net} \Delta \vec s[/itex]
    [itex]PE_i = mg \vec s_i[/itex]
    [itex]KE_f = \frac {1}{2}m {\vec v}^2[/itex]
    [itex]KE_i + PE_i = KE_f + PE_f \Rightarrow PE_i = KE_f \Rightarrow mg \vec s_i = \frac {1}{2}m {\vec v}^2 \Rightarrow \vec v = -\sqrt{2g \vec s_i}[/itex]

    (b) [itex]\vec F_{net2} = \vec F_f - \vec F_{net1} = ma \Rightarrow a = \frac {\vec F_f - \vec F_{net1}}{m}[/itex]
    [itex]v_f^2 = v_i^2 + 2 \vec a \vec s \Rightarrow \vec s = \frac {v_f^2 - v_i^2}{2 \vec a}[/itex]

    or

    [itex]F_f = ma \Rightarrow a = \frac {F_f}{m}[/itex]
    [itex]v_f^2 = v_i^2 + 2 \vec a \vec s \Rightarrow \vec s = \frac {v_i^2}{2 \vec a}[/itex]

    3. The attempt at a solution
    (a) [itex]\sum \vec F_x = F_{net1} = 400 N - (900 kg)(9.8 m/s^2)(sin 5.0^o) = -368.71 N[/itex]
    [itex]W_{net} = {\vec F}_{net} \Delta \vec s = (-368.71 N)(-50 m) = 18,435 J[/itex]
    [itex]\vec v = -\sqrt{2g \vec s_i} = -\sqrt{2(9.8 m/s^2)(50 m)} = -31.304 m/s \sim -31 m/s[/itex]

    (b) [itex]\vec v_i = -31.304 m/s[/itex]
    [itex]\vec v_f = 0 m/s[/itex]
    [itex]\vec F_f = 400 N[/itex]

    [itex]a = \frac {(400 N) - (-368.71 N)}{(900 kg)} = 0.8541 m/s^2[/itex]
    [itex]\vec s = \frac {(0)^2 - (-31.304 m/s)^2}{2(0.8541 m/s^2)} = -573.66 m \sim -570m[/itex]

    or

    [itex]F_f = ma \Rightarrow a = \frac {F_f}{m} = \frac {400 N}{900 kg} = 0.44444 m/s^2[/itex]
    [itex]\vec s = \frac {- (-31.304 m/s)^2}{2(0.44444 m/s^2)} = -1102.4 m \sim -1100m[/itex]

    (c) The distance will increase, because the greater mass will cause the force of gravity accelerating the car down the hill to increase. In turn, this will cause the net work, KE and velocity to increase.

    Is part a correct? Which way is correct for part b? Is anything else wrong?
    Thank-you
     
    Last edited: Nov 16, 2014
  2. jcsd
  3. Nov 16, 2014 #2
    Hi there. Part a is not correct. What is the height of the hill? The speed of the car at the bottom of the hill will be less than 31m/s because you left the work done by friction out of the equation. It should be PEi=KEf +Wf since the car was initially at rest. Think it in terms of energy.
     
    Last edited: Nov 16, 2014
  4. Nov 16, 2014 #3
    So with gravitational PE I use the vertical component of its displacement, but with work I use the displacement (which would be [itex]||\vec s|| sin \theta[/itex] and [itex]\Delta \vec s = s_f - s_i[/itex] respectively)?

    [itex]h = ||\vec s|| sin \theta = (50 m)sin(5.0^o) = 4.3577 m[/itex]

    Ok, so the car has some PE at the top, then it's converted to KE at the bottom, and in between there's a loss of energy due to friction. So:

    [itex]W_f = {\vec F}_f \Delta \vec s = (400 N)(-50 m) = 20,000 J[/itex]
    [itex]PE_i = KE_f + W_f \Rightarrow mgh = \frac {1}{2}m {\vec v_f}^2 + W_f \Rightarrow \vec v_f = -\sqrt {\frac {2(mgh - W_f)}{m}}[/itex]

    3. The attempt at a solution
    (a) [itex]\vec v_f = -\sqrt {\frac {2((900 kg)(9.8 m/s^2)(4.3577 m) - (20,000 J))}{(900 kg)}} = -7.9491 m/s \sim -7.9 m/s[/itex]

    (b) [itex]F_f = ma \Rightarrow a = \frac {F_f}{m} = \frac {400 N}{900 kg} = 0.44444 m/s^2[/itex]
    [itex]v_f^2 = v_i^2 + 2 \vec a \vec s \Rightarrow \vec s = \frac {v_i^2}{2 \vec a} = \frac {- (-7.9491 m/s)^2}{2(0.44444 m/s^2)} = -71.087 m \sim -71m[/itex]

    (c) The distance will increase, because the greater mass will cause the force of gravity accelerating the car down the hill to increase. In turn, this will cause the net work, KE and velocity to increase.

    Thank-you
     
    Last edited: Nov 16, 2014
  5. Nov 16, 2014 #4
    Exactly.

    Looks good now.
     
    Last edited: Nov 16, 2014
  6. Nov 16, 2014 #5
    This looks right as well.
    Nope, it won't increase. How is friction defined?
     
    Last edited: Nov 16, 2014
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