Answer:17.08 s
Explanation:
Given
distance between First and second Runner is 45.6 m
speed of first runner=3.1 m/s
speed of second runner=4.65 m/s
Distance between first runner and finish line is 250 m
Second runner need to run a distance of 250+45.6=295.6 m
Time required by second runner
time required by first runner to reach finish line
Thus second runner reach the finish line 80.64-63.56=17.08 s earlier
a) 2.64 s
We can solve this part of the problem by using the following SUVAT equation:
where
s is the displacement of the stone
u is the initial velocity
t is the time
a is the acceleration
We must be careful to the signs of s, u and a. Taking upward as positive direction, we have:
- s (displacement) negative, since it is downward: so s = -75.0 m
- u (initial velocity) positive, since it is upward: +15.5 m/s
- a (acceleration) negative, since it is downward: so a= g = -9.8 m/s^2 (acceleration of gravity)
Substituting into the equation,
Solving the equation, we have two solutions: t = -5.80 s and t = 2.84 s. Since the negative solution has no physical meaning, the stone reaches the bottom of the cliff 2.64 s later.
b) 10.4 m/s
The speed of the stone when it reaches the bottom of the cliff can be calculated by using the equation:
where again, we must be careful to the signs of the various quantities:
- u (initial velocity) positive, since it is upward: +15.5 m/s
- a (acceleration) negative, since it is downward: so a = g = -9.8 m/s^2
Substituting t = 2.64 s, we find the final velocity of the stone:
where the negative sign means that the velocity is downward: so the speed is 10.4 m/s.
c) 4.11 s
In this case, we can use again the equation:
where
s is the displacement of the package
u is the initial velocity
t is the time
a is the acceleration
We have:
s = -105 m (vertical displacement of the package, downward so negative)
u = +5.40 m/s (initial velocity of the package, which is the same as the helicopter, upward so positive)
a = g = -9.8 m/s^2
Substituting into the equation,
Which gives two solutions: t = -5.21 s and t = 4.11 s. Again, we discard the first solution since it is negative, so the package reaches the ground after
t = 4.11 seconds.
As we know by work energy theorem
total work done = change in kinetic energy
so here we can say that wok done on the box will be equal to the change in kinetic energy of the system
initial the box is at rest at position x = x1
so initial kinetic energy will be ZERO
at final position x = x2 final kinetic energy is given as
now work done is given as
so we can say
so above is the work done on the box to slide it from x1 to x2