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
Answer:
B) Force = 7.5, Time = 2 is equal to an impulse of 15 units
Answer: Hi!
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Hope this helps!
Explanation:
Mass of the wheel, m = 49 kg
Radius of the hoop, r = 0.73 m
Initial angular speed of the wheel,
Final angular speed of the wheel,
Time, t = 22 s
(a) If I is the moment of inertia of the hoop. It is equal to,
We know that the work done is equal to change in kinetic energy.
W = -1858.05 Joules
(b) Let P is the average power. It is given by :
P =84.45 watts
Hence, this is the required solution.