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sweet [91]
3 years ago
5

Imagine a system where a block rests on an inclined plane. The block is then given an initial push so that it starts sliding dow

n along the plane. After traveling some distance down the inclined plane it comes to rest due to frictional forces. In the simplest case, there are three forces acting upon the motion of the block—the initial push, gravity, and friction. Which of these statements about the work done by these three forces is true? The work done by the initial push is equal to the sum of the work done by friction and gravity. All three forces contribute an equal amount of work. The work done by friction is equal to the sum of the work done by gravity and the initial push. The work done by gravity is equal to the sum of the work done by friction and the initial push.
Physics
1 answer:
Helen [10]3 years ago
6 0

Answer:

statement - 'The work done by friction is equal to the sum of the work done by the gravity and the initial push' is correct.

Explanation:

The statement ''The work done by friction is equal to the sum of the work done by the gravity and the initial push" is correct.

The above statement is correct because, the initial push will tend to slide down the block thus the work done by the initial push will be in the downward direction. Also, the gravity always acts in the downward direction. thus, the work done done by the gravity will also be in the downward direction

here, the downward direction signifies the downward motion parallel to the inclined plane.

Now we know that the work done by the friction is against the direction of motion. Thus, the friction force will tend to move the block up parallel to the inclined plane.

Hence, for the block to stop sliding the the above statement should be true.

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castortr0y [4]

Answer:

W = 1222.4 J = 1.22 KJ

Explanation:

The work done on an object is the product of the force applied on it and the displacement it covers as a result of this force. It must be noted that the component of displacement in the direction of force should only be used. Hence, the work can be calculated as:

W = F d Cosθ

where,

W = Work Done = ?

F = Force Applied = 64 N

d = Distance Covered by Box = 19.1 m

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Therefore,

W = (64 N)(19.1 m)Cos 0°

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Answer:

frictional force = 0.52 N

Explanation:

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mass of turn table (M1) = 1.2 kg

diameter of shaft (D2) = 1.2 cm = 0.012 m

mass of shaft (M2) = 450 g = 0.45 kg

time (t) = 15 seconds

acceleration due to gravity (g) = 9.8 m/s^{2}

radius of turn table (R1) = 0.3 / 2 = 0.15 m

radius of shaft (R2) = 0.012 / 2 = 0.006 m

total moment of inertia (I) = moment of inertia of turn table + moment of inertia of shaft

I = 0.5(M1)(R1)^{2} + O.5 (M2)(R2)^{2}

I =  0.5(1.2)(0.15)^{2} + O.5 (0.45)(0.006)^{2}

I = 0.0135 + 0.0000081 = 0.0135081

ω₁ = 33 rpm = 33 x \frac{2π}{60} = 3.5 rad/s

α = -ω₁/t = -3.5 / 15 = -0.23 rad/s^{2}

torque = I x α

torque = 0.0135081 x (-0.23) = - 0.00311 N.m

torque = frictional force x R2

- 0.00311 = frictional force x 0.006

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 answer:
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