A 6.5 x 10^3 kg car accelerates from rest at the top ofa driveway that is sloped at an ...?

Question

angle of 15.9 deg with the horizontal. An average frictional force of 4.3 x 10^3 N impedes the car's motion so that the car's speed at the bottom of the driveway is 3.8 m/s. What is the length of the driveway? Answer in units of m.

Answer 1

using conservation of energy the kinetic energy at the bottom is
1/*m*v^2
The frictional force times the distance is the work done by friction
and the displacement in the vertical direction is the potential energy that gets converted to kinetic and friction

1/2*m*v^2+f*L=m*g*h
where h /L= sin(15.9)

Plug in the numbers and solve using algebra
(.5*6500*3.8*3.8)+4300*L=(6500*9.8*sin(15.9))*L
46930+4300L=17451*L
46930/(17451-4300)=L
=3.6m

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Answer 2

First enable's %. a coordinate equipment. enable the x axis be on the slope and the y axis would be perpendicular to the slope. Now the internet stress is 0 on the y axis because of the fact the vehicle isn't leaping or going everywhere vertically. Horizontally the internet stress is stress simply by gravity compelling the vehicle forward minus the friction stress. Now enable's examine the vector of the stress of gravity, shall we call it F_g for stress simply by gravity, and that's pointed vertically down. by ability of newtons 2d regulation we would desire to have F_g = mass*acceleration. right here the mass is the two.a million*10^3 kg automobile and acceleration is 9.8 m/s^2 that's the acceleration consistent of gravity. yet we are on a slope of 20 tiers and we will not use F_g immediately. somewhat we are able to ruin up F_g into an x component and a y component, which i will call F_gx and F_gy. we are actually not fascinated right here in F_gy when you consider that it is opposite to the traditional stress and all of us comprehend there is not any internet stress vertically (because of the fact the vehicle does not pass vertically). all of us comprehend there there's a internet stress in the horizontal course, so desire F_gx. by ability of a sprint geometry you are able to coach sin (20) = F_gx / F_g. Then F_g*sin(20) = F_gx. and F_g = 2100*9.8 so F_gx = 2100*9.8*sin(20). Now the internet stress horizontally = F_gx - F_friction = 2100*9.8*sin(20)- 4.0*10^3 = 3038.seventy seven N. when you consider that this stress is continuously utilized we've consistent acceleration. F = m*a -> F/m = a so a = 3038.seventy seven/ 2100 = a million.40 5 m/s^2. now we are able to apply our kinetic equations when you consider that we anticipate acceleration is continuing. a = delta v / delta t. v_i = 0 , v_f = unknown , and delta t is unknown besides. no issue, we are able to apply yet another kinetic equation delta x = a million/2 * a*t^2 + v_i * t . all of us comprehend delta x, and a, and v_i 5.8 = a million/2 * a million.40 5*t^2 t = sqrt( 2*5.8 / a million.40 5) = 2.eighty 3 seconds. now delta a = delta v / delta t -> delta v = a * delta t delta v = a million.40 5 * 2.eighty 3= 4.a million m/s observe : you're able to desire to get distinctive answer when you consider that i approximated

Answer 3

First we must determine how fast the car is accelerating,ac down this hill. For that we must determine the net Force,Fn taking frictional force,Ff and gravity,G into account along the plane.

Since Gravity acts downward we must take into account the hill so the gravitational force along the plane Fg = G * sin (theta)

and G = ma = 6.5 x 10^3 kg * 9.8 m/sec^2 = 6.37 x 10^4 N

Then Fn = Fg - Ff = 6.37 x10^4 * sin (15.9) - 4.3 x10^3 N = 1.315 x 10^4 N

then then the acceleration, ac = Fn / m = 1.315 x 10^4 / 6.5^3 = 2.023 m /sec^2

Now since the final velocity,vf is due to gravity and starts at rest, vo = 0.

So vf = at + vo = at (in this case)

so now we can solve for t = 3.8/2.023 = 1.88 secs

Then using the distance formula, with initial velocity,vo = 0 and initial distance, do = 0

d = 0.5*a*t^2 + vo *t + do = 0.5*2.02*(1.88)^2 = 3.58 m