A mysterious rocket-propelled object of mass 47.5kg is initially at rest in the middle of the horizontal, frictionless surface o
f an ice-covered lake. Then a force directed east and with magnitudeF(t)= (17.0N/s )t is applied.
How far does the object travel in the first 5.25s after the force is applied?
How far does the object travel in the first 5.25s after the force is applied?
1 answer:
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Answer:
(a) 91 kg (2 s.f.) (b) 22 m
Explanation:
Since it is stated that a constant horizontal force is applied to the block of ice, we know that the block of ice travels with a constant acceleration and but not with a constant velocity.
(a)
Subsequently,
*Note that the equations used above assume constant acceleration is being applied to the system. However, in the case of non-uniform motion, these equations will no longer be valid and in turn, calculus will be used to analyze such motions.
(b) To find the final velocity of the ice block at the end of the first 5 seconds,
According to Newton's First Law which states objects will remain at rest
or in uniform motion (moving at constant velocity) unless acted upon by
an external force. Hence, the block of ice by the end of the first 5
seconds, experiences no acceleration (a = 0) but travels with a constant
velocity of 4.4 .
Therefore, the ice block traveled 22 m in the next 5 seconds after the
worker stops pushing it.
Answer:
time of collision is
t = 0.395 s
so they will collide at height of 5.63 m from ground
Explanation:
initial speed of the ball when it is dropped down is
similarly initial speed of the object which is projected by spring is given as
now relative velocity of object with respect to ball
now since we know that both are moving under gravity so their relative acceleration is ZERO and the relative distance between them is 6.4 m
Now the height attained by the object in the same time is given as
so they will collide at height of 5.63 m from ground