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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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A block of ice(m = 14.0 kg) with an attached rope is at rest on a frictionless surface. You pull the block with a horizontal for
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b) The final speed of the block of ice is 9.8 meters per second.

Explanation:

a) We need to calculate the weight, normal force from the ground to the block and the pull force. By 3rd Newton's Law we know that normal force is the reaction of the weight of the block of ice on a horizontal.

The weight of the block (W), measured in newtons, is:

W = m\cdot g (1)

Where:

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g  - Gravitational acceleration, measured in meters per square second.

If we know that m = 14\,kg and g = 9.807\,\frac{m}{s^{2}}, the magnitudes of the weight and normal force of the block of ice are, respectively:

N = W = (14\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)

N = W = 137.298\,N

And the pull force is:

F_{pull} = 98\,N

The weight and the normal force of the block has a magnitude of 137.298 newtons and the pull force exerted on the block has a magnitude of 98 newtons.

b) Since the block of ice is on a frictionless surface and pull force is parallel to the direction of motion and uniform in time, we can apply the Impact Theorem, which states that:

m\cdot v_{o} +\Sigma F \cdot \Delta t = m\cdot v_{f} (2)

Where:

v_{o}, v_{f} - Initial and final speeds of the block, measured in meters per second.

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\Delta t - Impact time, measured in seconds.

Now we clear the final speed in (2):

v_{f} = v_{o}+\frac{\Sigma F\cdot \Delta t}{m}

If we know that v_{o} = 0\,\frac{m}{s}, m = 14\,kg, \Sigma F = 98\,N and \Delta t = 1.40\,s, then final speed of the ice block is:

v_{f} = 0\,\frac{m}{s}+\frac{(98\,N)\cdot (1.40\,s)}{14\,kg}

v_{f} = 9.8\,\frac{m}{s}

The final speed of the block of ice is 9.8 meters per second.

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