A motorcyclist is traveling at 51.9 mph on a flat stretch of highway during a sudden rainstorm. The rain has reduced the coeffic
ient of static friction between the motorcycle's tires and the road to 0.076 when the motorcyclist slams on the brakes. (a) What is the minimum distance required to bring the motorcycle to a complete stop? (b) What would be the stopping distance if it were not raining and the coefficient of static friction were 0.580?
2 answers:
Answer: a) 361.23m
b) 47.38m
Explanation:
Friction is opposite in direction to an applied force
Find the attached file for the solution
(b) The distance traveled with the friction of rain-free conditions is 47.38 m.
Answer:
Part (a) 361m
Part (b) 47.4m
This problem involves involves the nonconservative frictional force. To solve this problem. The energy approacb will be used. The energy conservation is used to relate the work done by the frictional force over the required distance to the kinetic energy of the motorcycle before the brakes are applied.
Explanation:
The equation is
K1 + U1 + Wf = K2 + U2
Where U standa for potential energy which in this case is zero as the height of the motorcycle does not change over the distance of the application of the brakes to the tyres. The work of friction is negative.
The full solution can be found in the attachment below.
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This question can be solved using the concept of friction energy.
The thermal energy change is b "258.4 J".
The change in thermal energy will be equal to the friction energy produced during the motion of the box.
where,
μ = coefficient of kinetic friction = 0.4
f = force applied = 38 N
d = distance traveled by the box = 17 m
Therefore,
<u>E = 258.4 J</u>
Learn more about friction energy here:
Answer:
3.95979 m/s
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration
Here
Initial velocity of the puck should be 3.95979 m/s
Answer:
The crate's coefficient of kinetic friction on the floor is 0.23.
Explanation:
Given that,
Mass of the crate, m = 300 kg
One worker pushes forward on the crate with a force of 390 N while the other pulls in the same direction with a force of 320 N using a rope connected to the crate.
The crate slides with a constant speed. It means that the net force acting on it is 0. Net force acting on it is given by :
So, the crate's coefficient of kinetic friction on the floor is 0.23.