consider the motion of the ball in vertical direction after it rolls off the table
v₀ = initial velocity of the ball in vertical direction = 0 m/s
a = acceleration = acceleration due to gravity = 9.8 m/s²
Y = vertical displacement = height of the table = 0.91 m
t = time spent in air by the ball
Using the kinematics equation
Y = v₀ t + (0.5) a t²
0.91 = 0 t + (0.5) (9.8) t²
t = 0.43 sec
v = final velocity in vertical direction just before it hits the ground
Using the kinematics equation
v = v₀ + at
v = 0 + (9.8) (0.43)
v = 4.21 m/s
The work done by the force in pulling the block all the way to the top of the ramp is 3.486 kJ.
<h3>What is work done?</h3>
Work done is equal to product of force applied and distance moved.
Work = Force x Distance
Given is a block with a weight of 620 N is pulled up at a constant speed on a very smooth ramp by a constant force. The angle of the ramp with respect to the horizontal is θ = 23.5° and the length of the ramp is l = 14.1 m.
From the Newton's law of motion,
ma =F-mg sinθ =0
So, the force F = mg sinθ
Plug the values, we get
F = 620N x sin 23.5°
F = 247.224 N
Work done by motor is W= F x d
The force is equal to the weight F = mg
So, W = 247.224 x 14.1
W = 3.486 kJ
Thus, the work done by the force in pulling the block all the way to the top of the ramp is 3.486 kJ.
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