An engine block of mass M is on the flatbed of a pickup truck that is traveling in a straight line down a level road with an initial speed of 25 m/s. The coefficient of static friction between the block and the bed is μs = 0.566. Find the minimum distance in which the truck can come to a stop without the engine block sliding toward the cab in meters.
s =v ² /2•a = v ² /2•μ•g = 25²/2•0.566•9.8 =56.34 m
To find the minimum distance in which the truck can come to a stop without the engine block sliding toward the cab, we need to consider the forces acting on the block and the maximum friction force that can prevent the block from sliding.
Let's analyze the forces acting on the engine block:
1. Weight (mg): The weight of the block acts vertically downward and has a magnitude of Fg = mg, where m is the mass of the block and g is the acceleration due to gravity.
2. Normal force (N): The normal force is the force exerted by the flatbed of the truck on the block and acts vertically upward. Since the block is on a level road, the normal force has the same magnitude as the weight, i.e., N = mg.
3. Friction force (f): The friction force opposes the motion and can prevent the block from sliding. The maximum friction force is given by the equation f = μsN, where μs is the coefficient of static friction and N is the normal force.
Now, let's find the maximum friction force:
f = μsN
= μsmg
To stop the block without sliding, the maximum friction force (f) should be equal to or greater than the force that would cause the block to slide downward or forward. This force is the component of the weight of the block parallel to the bed, given by Fslide = mgsinθ, where θ is the angle between the direction of motion and the vertical.
For example, if the block were to slide downward:
f ≥ Fslide
μsmg ≥ mgsinθ
The mass cancels out:
μsg ≥ gsinθ
Since g is positive, we can divide both sides by g:
μs ≥ sinθ
Now, we can find the minimum distance by considering the acceleration required to stop the truck:
The net force acting on the block is given by:
Net force (Fnet) = friction force (f)
Using Newton's second law, Fnet = f = ma, where a is the acceleration.
Also, the friction force can be written as f = μsN = μsmg.
Setting Fnet equal to f, we have:
ma = μsmg
The mass cancels out:
a = μsg
To stop the truck, the final velocity will be zero (v = 0), and the initial velocity is given as 25 m/s.
Using the kinematic equation v^2 = u^2 + 2as, where u is the initial velocity, v is the final velocity, a is the acceleration, and s is the distance:
0 = (25 m/s)^2 + 2(-a)s
Substituting the value of a, we have:
0 = (25 m/s)^2 + 2(-μsg)s
0 = (25 m/s)^2 - 2μsgs
Rearranging the equation, we get:
2μsgs = (25 m/s)^2
s = [(25 m/s)^2] / (2μsg)
Substituting the given value of the coefficient of static friction (μs = 0.566), and the acceleration due to gravity (g = 9.8 m/s^2), we can calculate the minimum distance (s) in meters.
To find the minimum distance in which the truck can come to a stop without the engine block sliding toward the cab, we need to consider the forces acting on the engine block.
1. The gravitational force (Fg): The weight of the engine block can be calculated using the formula Fg = mg, where m is the mass of the block and g is the acceleration due to gravity (approximately 9.8 m/s^2).
2. The frictional force (Ff): The frictional force between two surfaces can be calculated using the formula Ff = μN, where μ is the coefficient of friction and N is the normal force. In this case, the normal force is equal to the weight of the engine block (N = Fg).
3. The net force (Fnet): The net force acting on the block is the difference between the applied force (ma, where a is the acceleration) and the frictional force (Ff).
For the block to remain stationary, the net force should be zero (Fnet = 0), and thus the applied force should be equal to the frictional force (Ff).
Now, let's calculate the minimum distance required:
1. First, find the frictional force (Ff) using the formula Ff = μN, where N = Fg:
Ff = μs * Fg
2. Set the frictional force equal to the applied force (ma):
Ff = ma
3. Equate the equations from step 1 and step 2:
μs * Fg = ma
4. Substitute the values and solve for acceleration (a):
μs * mg = ma
a = μs * g
5. Use the kinematic equation v^2 = u^2 + 2as, where u is the initial velocity, v is the final velocity (0 m/s as the block comes to a stop), and s is the distance:
0 = (25 m/s)^2 + 2 * a * s
6. Solve for s (the minimum distance):
s = -u^2 / (2 * a)
s = -(25 m/s)^2 / (2 * a)
Now, plug in the known values to calculate the minimum distance.