Aaron's car is moving at speed of 30 m/s
His reaction time is given as 0.7 s
but when he is tired the reaction time is doubled
Now we need to find the distance covered by his car when he is tired during the time when he react to apply brakes
So here since during this time speed is given as constant so we can say that distance covered can be product of speed and time
So here we can use
So the car will move to 42 m during the time when he apply brakes
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