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polet [3.4K]
3 years ago
10

A 5.0-kg box rests on a horizontal surface. The coefficient of kinetic friction between the box and surface is ????K=0.50. A hor

izontal force pulls the box at constant velocity for 10 cm. Find the work done by (a) the applied horizontal force, (b) the frictional force, and (c) the net force.
Physics
1 answer:
blagie [28]3 years ago
3 0

Answer:

a)W= 1 J

b)W= 1 J

c)W= 0 J

Explanation:

Given that

m = 5 kg

Coefficient of kinetic friction,K=0.5

d= 10 cm

We know that if velocity is constant it means that the acceleration of the system is zero.Or we can say that all forces are balance in the system.

We know that work is the dot product of force and displacement

W= F.d

The friction force on the box

Fr= K m g

Fr= 0.5 x 5 x 10 = 25 N                      ( take g =10 m/s²)

Fr=25 N

a)

Acceleration a= 0

So horizontal force F = Fr

W = Fr.d

W=10 x 0.1 J

W= 1 J

b)

The work done by friction force

W= 1 J

c)

The net force on the system is zero because acceleration is zero.

F= 0

So

W= 0 J

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sammy [17]

Answer: 3.27kg

Explanation:

Inertia can be said to be the resistance of an object to a change in its motion. This includes a change in its direction. An object will stay still or keep moving at the same speed and in a straight line, except it is acted upon by an unbalanced external force.

Given F = 30N

r = 0.14m

a = 130rad/s²

Then, T = Fr

T = 30*0.14

T = 4.2Nm

Also, T = inertia * acceleration

Inertia = 4.2/a

Inertia = 4.2/130

Inertia = 0.032

Also, inertia = mr²/2

0.032 = m * (0.14²)/2

0.032 = m * 0.0098

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3 years ago
A duck swimming on the surface of a pond has an
Gemiola [76]

From the given information:

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However, we will have to compute the initial velocity and the acceleration of the duck in their vector forms.

<h3>In vector form;</h3>

The initial velocity is:

\mathbf{u ^{\to} = 0.7 m/s ( -cos 25^0 \hat x + sin 25^0 \hat y ) \ m/s}

The acceleration is:

\mathbf{a ^{\to} = 0.5 m/s ( cos 41^0 \hat x - sin 41^0 \hat y ) \ m/s^2}

The objective of this question is to determine the speed of the duck at a certain time. Since it is not given, let's assume we are to determine the Duck speed after 4 seconds of accelerating;

Then, it implies that time (t) =  4 seconds.

Using the first equation of motion:

v^{\to} = u ^{\to} + a^{\to} t

Then, we can replace their values into the equation of motion in order to determine the speed:

\mathbf{v^{\to} =\Big(0.7 ( -cos 25^0 \hat x + sin 25^0 \hat y )+4 \times 0.5 ( cos 41^0 \hat x - sin 41^0 \hat y )\Big)}

\mathbf{v^{\to} =\Big(0.7 ( -cos 25^0 \hat x + sin 25^0 \hat y )+2.0 ( cos 41^0 \hat x - sin 41^0 \hat y )\Big)}

\mathbf{v^{\to} =\Big( ( -0.7 cos 25^0 \hat x + 0.7 sin 25^0 \hat y )+( 2.0cos 41^0 \hat x - 2.0sin 41^0 \hat y )\Big)}

Collect like terms:

\mathbf{v^{\to} =\Big( (2.0cos 41^0 -0.7 cos 25^0   )\hat x+(  0.7 sin 25^0 - 2.0sin 41^0 )\Big)\hat y}

\mathbf{v^{\to} =0.87500   \hat x- 1.01629 \hat y}

Thus, the magnitude is:

\mathbf{v^{\to} =\sqrt{(0.87500 )^2 +( 1.01629 )^2}}

\mathbf{v^{\to} =\sqrt{0.76563 +1.03285}}

\mathbf{v^{\to} =\sqrt{1.79848}}

\mathbf{v^{\to} =1.34 \ m/s}

Therefore, we can conclude that the speed of the duck after 4 seconds is 1.34 m/s

Learn more about vectors here:

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lawyer [7]

Answer:

A is the answer!

Explanation:

Edge 2021

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Dima020 [189]

Initial Velocity is the velocity at time interval t = 0 and it is represented by u. It is the velocity at which the motion starts.

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