help me with my friction take home test and i will love you =)?

Question

ok here it is ..
a car is going 40m/s sees a mosse in the roadway ahead. (a) if the coefficent of friction for rubber on asphalt is .95 how far down the road must the moose be to avoid being hit? (b) if the moose is 35 meter down the road, what will be the speed of impact?

thank you soo much

Answer 1

for both you use the equation:

vfinal^2 = vinitial^2 +2ad where a is acceleration and d is distance. For both parts, we need to know the acceleration and that is contained in the coefficient of friction. We know that the frictional force between the road and the car generates a force equal to mu mg, where mu is the coefficient of friction, m is the mass of the car and g is the accel due to gravity. We also know that forces cause accelerations according to newton's second law, F=ma, so equating the forces, we have that:

-mu mg = ma => a = -mu g = -0.95 x 9.81 m/s/s = -9.31 m/s/s in this case. The minus sign occurs because the speed is slowing down, as we expect with friction.

back to part a: the stopping distance occurs when vfinal=0, so:

vf^2 = vi^2 +2ad; vi=40, vf = 0, so:

d=vi^2/2a = 40^2/(2x9.31) = 85.9 m

part b: if the moose is 35 m down the road:

vf^2 = vi^2 +2ad => vf^2 = 40^2 -2(9.31)(35) or

vf=30.7 m/s, or you have one dead moose.

Answer 2

If the car has mass m, you can start by calculating its initial kinetic energy KE_0 = 0.5*mv^2, where v is the given speed of 40 m/s. The force of friction would be F_f = umg, where u is the coefficient of friction (given as 0.95) and g is the acceleration due to gravity (9.8 m/s^2). The work done by friction is W_f = (F_f)*x, where x is the distance along which it acts. The work done by friction must be equal to the initial kinetic energy in order for the final velocity to be zero. So say KE_0 = W_f ==> 0.5*mv^2 = umgx, where x is your only unknown.

If the moose is 35 m down the road, the work done by friction is W_f = (F_f)*x, where x is now the given 35 m. The kinetic energy of the car becomes KE = KE_0 - W_f, and then use the formula KE = 0.5*mv^2 to solve for v, the velocity of impact.

I chose to use energy balance. However, you can also use v^2 = (v_0)^2 + 2ax, where v is final velocity, v_0 is initial velocity, a is acceleration, and x is distance. The acceleration in this case is ug, due to the friction relationship.

Answer 3

Friction=µR
F=ma
µ(mg)=ma
-0.95(9.8)=a
acceleration = -9.31 m/s²

V²=2as+U²
V is final velocity
a=-9.31
s=??
U intial velocity
(0)²=2(-9.31)(s)+(40)²
s=85.92910849 m

V²=2as+U²
V²=2(-9.31)(35)+(40)²
V²=948.3 m²/s²
V=30.79448002 m/s

Answer 4

40 * 0.95 = 38
1.05m/s