It has potential energy because of its position.
Answer:
acceleration = -0.042 m/s²
velocity at beginning = 14.167 m/s
velocity at end = 5.7183 m/s
Explanation:
given data
distance d1 = 1 km
distance d2 = 2 km
time t1 = 80 s
time t2 = 120 s + 80s = 200 s
to find out
acceleration and velocity at beginning and end
solution
we apply here law of motion that is
d = vt + 1/2×at²
put value
1000 = v(80) + 1/2×a(80)² ........................1
and
2000 = v(200) + 1/2×a(200)² ........................2
so from equation 1 and 2 we get a and v
a = -0.042 m/s² and
v = 14.167 m/s
so by kinematic final velocity will be
V² = v² + 2ad
V² = (14.167)² + 2×(-0.042)×(2000)
V² = 32.70
V = 5.7183 m/s
so
acceleration = -0.042 m/s²
velocity at beginning = 14.167 m/s
velocity at end = 5.7183 m/s
Answer:
The minumum speed the pail must have at its highest point if no water is to spill from it
= 2.64 m/s
Explanation:
Working with the forces acting on the water in the pail at any point.
The weight of water is always directed downwards.
The normal force exerted on the water by the pail is always directed towards the centre of the circle of the circular motion.
And the centripetal force, which keeps the system in its circular motion, is the net force as a result of those two previously mentioned force.
At the highest point of the motion, the top of the vertical circle, the weight and the normal force on the water are both directed downwards.
Net force = W + (normal force)
But the speed of this motion can be lowered enough to a point where the normal force becomes zero at the moment the pail reaches the highest point of its motion. Any speed lower than this value would result in the water spilling out of the pail. The water would not be able to resist the force of gravity.
At this point of minimum velocity,
Normal force = 0
Net force = W
Net force = centripetal force = (mv²/r)
W = mg
(mv²/r) = mg
r = 0.710 m
g = 9.8 m/s²
v² = gr = 9.8 × 0.71 = 6.958
v = √(6.958) = 2.64 m/s
Hope this Helps!!!
