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2 years ago

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How to prove3 equations of motion graphically

1 answer:

Softa [21]2 years ago

6 0

Consider an object is moving with a uniform acceleration “a” along a straight line. The initial and final velocities of the object at time t = 0 and t = t are u and v respectively. During time t, let s be the total distance travelled by the object.

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2 years ago

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Are collision more likely to happen between stars planets or galaxies

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2 years ago

A 2.0 kg block is pushed 1.0 m at a constant

svet-max [94.6K]

1) Work done by the force: 8.3 J

2) Work done by gravity: -19.6 J

3) Normal force: 16.3 N

Explanation:

1)

The work done by a force pushing an object is given by

$W=Fd cos \alpha$

where

F is the magnitude of the force

d is the displacement of the object

$\alpha$ is the angle between the direction of the force and of the displacement

First of all, we need to find the magnitude of the force F. Since the block is moving vertically at constant velocity (= zero acceleration), the equation of motion of the block along the vertical direction is:

$F sin \theta - \mu N - mg = 0$ (1)

Where

$\theta=27.0^{\circ}$

$\mu N$ is the force of friction, where

$\mu=0.40$ is the coefficient of friction

N is the normal reaction of the wall

(mg) is the weight of the block, where

m =2.0 kg is the mass of the block

$g=9.81 m/s^2$ is the acceleration of gravity

Along the horizontal direction, the equation of motion is:

$F cos \theta = N$ (2)

Substituting (2) into (1),

$F sin \theta - \mu F cos \theta - mg =0$

And solving for F,

$F(sin \theta - \mu cos \theta) = mg\\F=\frac{mg}{sin \theta - \mu cos \theta}=\frac{(2.0)(9.81)}{sin 27^{\circ} - 0.40 cos 27^{\circ}}=18.3 N$

Now we can calculate the work done by the force. Here we have

d = 1.0 m is the displacement

$\alpha = 90^{\circ} - 27^{\circ} = 63^{\circ}$ is the angle between the direction of the force and the displacement

Substituting,

$W=(18.3)(1.0)(cos 63^{\circ})=8.3 J$

2)

The work done by gravity is equal to:

$W_g = mg d cos \alpha$

where

m = 2.0 kg is the mass of the block

$g=9.81 m/s^2$

d = 1.0 m is the displacement

$\alpha=180^{\circ}$ , since the direction of gravity is downward while the displacement of the block is upward

Substituting,

$W=(2.0)(9.81)(1.0)(cos 180^{\circ})=-19.6 J$

3)

Equation (2) found in part 1) tells that

$N=F cos \theta$

which means that the normal force between the wall and the block is equal to the horizontal component of the pushing force on the block.

Here we have:

F = 18.3 N is the magnitude of the pushing force

$\theta=27.0^{\circ}$ is the angle of the force with the horizontal

Substituting, we find the normal force:

$N=(18.3)(cos 27^{\circ})=16.3 N$

Learn more about forces and acceleration:

brainly.com/question/11411375

brainly.com/question/1971321

brainly.com/question/2286502

brainly.com/question/2562700

#LearnwithBrainly

8 0

2 years ago

A 90.0-kg mail bag hangs by a vertical rope 3.5 m long. A postal worker then displaces the bag to a position 2.0 m sideways from

asambeis [7]

Answer:

615 N

Explanation:

If θ is the angle from vertical and T is the rope tension

θ = arcsin (2.0 / 3.5) = 34.85°

Summing vertical forces to zero

Tcosθ - mg = 0

T = 90.0(9.81) / cos34.85 = 1,075.85 N

Summing horizontal forces to zero

F - Tsinθ = 0

F = 1075.85sin34.85 = 1075.85(2.0/3.5) = 614.772... ≈ 615 N

3 0

2 years ago