A milk truck carries milk with density 64.6 lb/ft3 in a horizontal cylindrical tank with diameter 12 ft. (a) Find the force exer
1 answer:
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Answer:
The total distance is 381.5 [m]
Explanation:
In order to solve this problem we must use the expressions of kinematics. The clue to solve this problem is that the motorcyclist starts from rest, i.e. its initial speed is zero.
where:
Vf = final velocity [m/s]
Vo = initial velocity = 0
a = acceleration = 2 [m/s²]
t = time = 7 [s]
Vf = 0 + (2*7)
Vf = 14 [m/s]
With this velocity, we can calculate the displacement using the following expression.
where
x = distance traveled [m]
14² = 0 + (2*7*x)
x = 196/(14)
x = 14 [m]
Note: The positive sign in the equations is because the car is accelerating, it means its velocity is increasing.
The other important clue to solve this problem in the second part is that the final velocity is now the initial velocity.
We must calculate the final velocity.
Vf = final velocity [m/s]
Vi = initial velocity = 14 [m/s]
a = desacceleration = 4 [m/s²]
t = time = 8 [s]
Vf = 24 + (4*8)
Vf = 56 [m/s]
With this velocity, we can calculate the displacement using the following expression.
where
x = distance traveled [m]
56² = 14² + (2*4*x)
x = 2940/(8)
x = 367.5 [m]
Note: The positive sign in the equations is because the car is accelerating, it means its velocity is increasing.
Therefore the total distance is Xt = 14 + 367.5 = 381.5 [m].
Answer:
a. 79.1 N
b. 344 J
c. 344 J
d. 0 J
e. 0 J
Explanation:
a. Since the crate has a constant velocity, its net force must be 0 according to Newton's 1st law. The push force by the worker must be equal to the friction force
on the crate, which is the product of friction coefficient μ and normal force N:
Let g = 9.81 m/s2
b. The work is done on the crate by this force is the product of its force and the distance traveled s = 4.35
c. The work is done on the crate by friction force is also the product of friction force and the distance traveled s = 4.35
This work is negative because the friction vector is in the opposite direction with the distance vector
d. As both the normal force and gravity are perpendicular to the distance vector, the work done by those forces is 0. In other words, these forces do not make any work.
e. The total work done on the crate would be sum of the work done by the pushing force and the work done by friction