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2 years ago

A 16.7 kg block is dragged over a rough, horizontal surface by a constant force of 132 N acting

Physics

1 answer:

Alisiya [41]2 years ago

6 0

The magnitude of the work done by the force of friction is 1045.68 joules

Explanation:

Work done = force × distance

Work done by friction force (W) = friction force (F) × distance (d)
Friction force (F) = coefficient of kinetic friction (μ) × normal reaction force (R)

A 16.7 kg block is dragged over a rough, horizontal surface by a constant force of 132 N acting at an angle of 25.8. above the horizontal. The block is displaced 57,8 m, and the coefficient of kinetic friction is 0.229

To find the normal reaction force of the block distribute The constant force into two component, horizontal component of 132 cos(25.8°) and vertical component 132 sin(25.8°), Then equate ∑vertical forces by 0 to find R

The weight of the block = mg, where m is its mass and g is 9.8 m/s²

→ ∑Vertical forces = R + 132 sin(25.8°) - mg

→ ∑Vertical forces = R + 132 sin(25.8°) - (16.7)(9.8)

→ ∑Vertical forces = R - 106.2

Equate it by 0

→ R - 106.2 = 0 ⇒ add 106.2 to both sides

→ R = 106.2 N

Now let us find the friction force

→ F = μ R

→ μ = 0.229

→ F = 0.229 (106.2) = 24.3198 N

Let us calculate the work don by the force of friction

→ W = F × d

→ d = 57.8 meters

→ W = 24.3198 × 57.8 = 1045.68 joules

The magnitude of the work done by the force of friction is 1045.68 joules

Learn more:

You can learn more about force of friction in brainly.com/question/6217246

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Explanation: This question involves the concept of Latent Heat and specific heat capacities of water in solid phase.

Latent heat of fusion is the total amount of heat rejected from the unit mass of water at 0 degree Celsius to convert completely into ice of 0 degree Celsius (and the heat required for vice-versa process).

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Asked:

1. Time require for the ice to start melting.

2. Time required to raise the temperature above freezing point.

Sol.: 1.

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Using above equation we find the total heat required to bring the ice from -16.6°C to 0°C.

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Now, we require 19347.3 joules of heat to bring the ice to 0°C and then on further addition of heat it starts melting.

∴The time required before the ice starts to melt is the time required to bring the ice to 0°C.

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Sol.: 2.

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Now comes the concept of Latent heat into the play, the temperature does not starts rising for the ice as soon as it reaches at 0°C it takes significant amount of time to raise the temperature because the heat energy is being used to convert the phase of the water molecules from solid to liquid.

From the above solution we have concluded that 23.59 minutes is required for the given ice to come to 0°C, now we need some extra amount of energy to convert this ice to liquid water of 0°C.

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