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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Answer:
712.5 kg m/s
Explanation:
Work out the total momentum before the event (before the collision):
p = m × v
Massof Deon = 95kg
Velocity =7.5m/s
Mass of Chuck = 120kg
Velocity = 0m/s
Momentum of Deon before = 95 × 7.5 = 712.5 kg m/s
Momentum of Chuck before = 120 × 0 = 0 kg m/s
Total momentum before = 712.5+ 0 = 712.5 kg m/s
Working out the total momentum after the event (after the collision):
Because momentum is conserved, total momentum afterwards = 712.5 kg m/s
Answer:
Zero
Explanation:
As we know that the force and the motion direction should always be perpendicular to each other due to which the work is done by static friction be zero
Therefore
F.dcos(theta) = F.d cos(90) = 0
Hence, the work done by static friction is zero
Therefore the same is to be considered
