A 900 kg motorcycle traveling at 20 m/s slides to a stop after applying the brak

Question

A 900 kg motorcycle traveling at 20 m/s slides to a stop after applying the brakes. If the surface of the road has a coefficient of kinetic friction of 0.40, how far does the motorcycle skid? A shuffleboard puck having an initial speed of 5 m/s slides 8.0 m before coming to rest. What is the coefficient of kinetic friction between the puck and the surface? While redecorating his apartment, Bob slowly pushes an 82 kg cabinet across a wooden floor, which resists the motion with a force of friction of 320 N. What is the coefficient sliding friction between the cabinet and the floor? A 2 kg crate is shot up a 20 degree incline at 30 m/s. How far up the incline will it slide? How long will it take to do this? A 2 kg crate is shot up a 20 degree incline at 30 m/s. If the incline's surface has a coefficient of kinetic friction of 0.30, How far up the incline will it slide? How long will it take to do this? In the soccer finals, Elizabeth kicks a 0.6 kg soccer ball with a force of 80 N. How much does she accelerate the soccer ball from rest in the process? A 2 kg crate with initial speed of 3 m/s slides 2 m to a stop. What is the coefficient of kinetic friction? What force would be needed to keep the crate moving at a constant speed? A box starting from rest slides down a 30degree incline. What is the box's acceleration? What is the speed of the box after it has traveled 18 m? A box starting from rest slides down a 30degree incline. After traveling 18 m the box has a speed of 3 m/s. What is the box's acceleration? What is the coefficient of kinetic friction between the box and the incline? A hockey puck leaves a player's stick with a speed of 10 m/s and slides 40 m before coming to rest. What is the coefficient of friction between the puck and the ice?

Explanation / Answer

At one time, I can solve maximum of 4 questions.

1) Using Newton's 2nd Law of Motion, F = ma
where
F = net force acting on the puck = -µmg
µ = coefficient of friction
m = mass of the puck
g = acceleration due to gravity = 9.8 m/sec^2 (constant)
and since F = ma, then
-µmg = ma ========> a = µg = 0.4*9.8 = 3.92 m/s^2

Now Vf^2 - Vo^2 = 2as

where
Vf = final velocity = 0 (when the puck stops)
Vo = initial velocity = 20 m/sec. (given)
a = acceleration = -3.92 m/s^2
s = stopping distance = ?

s = Vo^2/2a = 400/(2*3.92) = 51.02 m

2) First working formula is Vf^2 - Vo^2 = 2as
where
Vf = final velocity = 0 (when the puck stops)
Vo = initial velocity = 5 m/sec. (given)
a = acceleration
s = stopping distance = 8 m (given)

Substituting appropriate values,
0 - 5^2 = 2(a)(8)
a = -1.56 m/sec^2
NOTE the negative sign attached to the acceleration. It simply means that the puck was slowing down due to the friction between it and the surface.

Using Newton's 2nd Law of Motion, F = ma
where
F = net force acting on the puck = -µmg
µ = coefficient of friction
m = mass of the puck
g = acceleration due to gravity = 9.8 m/sec^2 (constant)
and since F = ma, then
-µmg = ma
Again, note the negative sign attached to "F = - µmg" as this implies that the direction of the frictional force is opposite that of the puck's motion.
Since "m" appears on both sides of the equation, it will simply cancel out and the above becomes
- µ(9.8) = -1.56
and solving for µ
µ = 0.159

3) Frictional Force = f = µN = µmg =========> µ = f/mg = 320/(82*9.8) = 0.398

4) there are 2 forces which cause the crate to decelerate, force parallel and friction force.

Ff = 0.3 * 2 * 9.8 * cos 20 = 5.88 * cos 20
Total force = 19.6 * sin 20 + 5.88 * cos 20
To determine the deceleration divide by 2.

a = 9.8 * sin 20 + 2.94 * cos 20
This is approximately -6.11 m/s^2.
To determine the distance it slides, use the following equation.

vf^2 = vi^2 + 2 * a * d, vf = 0
0 = 30^2 + 2 * -(9.8 * sin 20 + 2.94 * cos 20) * d
d = 900 ÷ [2 * -(9.8 * sin 20 + 2.94 * cos 20)]
This is approximately 73.6 meters.

To determine the time, use the following equation.

vf = vi + a * t
0 = 30 – (9.8 * sin 20 + 2.94 * cos 20) * t
t = 30 ÷ (9.8 * sin 20 + 2.94 * cos 20)
The time is approximately 4.91 seconds.

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