The floor of a railroad flatcar is loaded with loose crates having a coefficient of static friction 0.25 of with the floor. If the train is initially moving at a speed of 48 km/h, in how short a distance can the train be stopped at constant acceleration without causing the crates to slide over the floor?
The distance is 36.23 m
Coefficient of static friction,
Initial speed of the train,
Final speed of the train,
The problem is based on Newton’s second law of motion which states that the rate of change of momentum of a body is equal in both magnitude and direction of the force acting on it. Use the Newton's 2nd law of motion to find the acceleration of the train and then by using the Kinematic equations of motion, find the distance travelled by train.
The frictional force acting on the train is given by,
By using Newton's 2nd law along the horizontal direction,
This is the acceleration of the train.
Now, by using the Kinematic equation of motion,
Therefore, the distance is 36.23 m.
A four-person bobsled comes down a straightaway at the start of a bobsled run. The straightaway is 80.0 m long and is inclined at a constant angle of with the horizontal. Assume that the combined effects of friction and air drag produce on the bobsled a constant force of 62.0 N that acts parallel to the incline and up the incline. Answer the following questions to three significant digits.
(a) If the speed of the bobsled at the start of the run is 6.20 m/s, how long does the bobsled take to come down the straightaway?
(b) Suppose the crew is able to reduce the effects of friction and air drag to 42.0 N. For the same initial velocity, how long does the bobsled now take to come down the straightaway?
Body A in Fig. 6-33 weighs , and body B weighs . The coefficients of friction between A and the incline are and . Angle is . Let the positive direction of an axis be up the incline. In unit-vector notation, what is the acceleration of A if A is initially (a) at rest, (b) moving up the incline, and (c) moving down the incline?
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