You toss a ball straight up in the air. Immediately after you let go of it, what force or forces are acting on the ball
1 answer:
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Answer: = 16.49N
Explanation: The object is placed on an inclined plane at an angle of 37° thus making it weight have two component,
= horizontal component of the weight = mgsinФ
= vertical component of weight = mgcosФ
Due to the way the object is positioned, the horizontal component of force will accelerate the object thus acting as an applied force.
by using newton's law of motion, we have that
mgsinФ - = ma
where m = mass of object=5kg
a = acceleration= unknown
Ф = angle of inclination = 37°
g = acceleration due to gravity = 9.8
= frictional force = unknown
we need to first get the acceleration before the frictional force which is gotten by using the equation below
where v = final velocity = 2m/s
u = initial velocity = 0m/s (because the object started from rest)
a= unknown
S= distance covered = length of plane = 5m
we slot in a into the equation below to get frictional force
mgsinФ - = ma
3 * 9.8 * sin 37 - = 3* 0.4
17.9633 - = 1.2
= 17.9633 - 1.2
= 16.49N