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mezya [45]
2 years ago
14

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.
Calculate the work done by the force in pulling the block all the way to the top of the ramp. (Neglect friction.)

Physics
1 answer:
GuDViN [60]2 years ago
7 0

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.

Learn more about Work done

brainly.com/question/13662169

#SPJ1

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Answer:

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y=y_{o} +v_{o} *t-\frac{1}{2} *g*t^{2} \\y=63.72 +22.05 *t-\frac{1}{2} *(9.81)*t^{2} \\\\y=63.72 +22.05 *t-4.905*t^{2} \\

As we can see the equation is based on Time (t).

Now we can establish with the conditions of the ball launched by David a new equation for y (elevation) in function of t, then we match these equations and find time t

y=y_{o} +v_{o} *t+\frac{1}{2} *g*t^{2} \\where:\\v_{o} =55.56[m/s] = initial velocity\\y_{o} =0[m]\\now replacing\\63.72 +22.05 *t-(4.905)*t^{2} =0 +55.56 *t-(4.905)*t^{2} \\63.72 +22.05 *t =0 +55.56 *t\\63.72 = 33.51*t\\t=1.9[s]

Then the time when both balls are going to be the same height will be when 2 [s] plus 1.9 [s] have passed after David throws the ball.

Time = 2 + 1.9 = 3.9[s]

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