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oee [108]
3 years ago
9

An object of mass 6 kg. is resting on a horizontal surface. A horizontal force

Physics
1 answer:
son4ous [18]3 years ago
7 0

Answer:

a) The work done by the applied force is 1500 joules.

b) The kinetic energy of the block after 10 seconds is 1200 joules.

c) The magnitude of the force of friction is 3 newtons and its direction is against motion.

d) 300 joules of energy are lost during motion.

Explanation:

a) Since the object has a constant mass, on which a constant horizontal force is exerted. The work done by the force (W), measured in joules, is defined by the following expression:

W = F\cdot \Delta x (1)

Where:

F - Force, measured in newtons.

\Delta x - Distance, measured in meters.

If we know that F = 15\,N and \Delta x = 100\,m, then the work done by the force exerted on the object is:

W = (15\,N)\cdot (100\,m)

W = 1500\,J

The work done by the applied force is 1500 joules.

b) At first we need to calculate the net acceleration of the object (a), measured in meters per square second. By assuming a constant acceleration, we use the following kinematic formula:

\Delta x = v_{o}\cdot t +\frac{1}{2}\cdot a\cdot t^{2} (2)

Where v_{o} is the initial velocity of the object, measured in meters per second.

We clear the acceleration within the equation above:

\frac{1}{2}\cdot a \cdot t^{2} =  \Delta x-v_{o}\cdot t

a = \frac{2\cdot (\Delta x - v_{o}\cdot t)}{t^{2}}

If we know that \Delta x = 100\,m, v_{o} = 0\,\frac{m}{s} and t = 10\,s, then the net acceleration experimented by the object is:

a = \frac{2\cdot \left[100\,m-\left(0\,\frac{m}{s} \right)\cdot (10\,s)\right]}{(10\,s)^{2}}

a = 2\,\frac{m}{s^{2}}

By the 2nd Newton's Law, we construct the following equation of equilibrium under the consideration of a friction force acting against the motion of the object:

\Sigma F = F - f = m\cdot a (3)

Where:

F - External force exerted on the object, measured in newtons.

f - Kinetic friction force, measured in newtons.

If we know that F = 15\,N, m = 6\,kg and a = 2\,\frac{m}{s^{2}}, the kinetic friction force is:

f = F-m\cdot a

f = 15\,N-(6\,kg)\cdot \left(2\,\frac{m}{s^{2}} \right)

f = 3\,N

The work done by friction (W'), measured in joules, is:

W' = f\cdot \Delta x (4)

W' = (3\,N) \cdot (100\,m)

W' = 300\,J

And the net work experimented by the object is:

\Delta W = 1500\,J - 300\,J

\Delta W = 1200\,J

By the Work-Energy Theorem we understand that change in translational kinetic energy (\Delta K), measured in joules, is equal to the change in net work. That is:

\Delta K = \Delta W (5)

If we know that \Delta W = 1200\,J, then the change in translational kinetic energy is:

\Delta K = 1200\,J

The kinetic energy of the block after 10 seconds is 1200 joules.

c) The magnitude of the force of friction is 3 newtons and its direction is against motion.

d) The energy lost by the object is equal to the work done by the force of friction. Therefore, 300 joules of energy are lost during motion.

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<h3>What is pressure?</h3>

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The four different types of pressure are absolute, atmospheric, differential, and gauge pressure. Have you ever noticed that when you use a straw to drink something, the air actually gets suked out? In reality, you're applying "Pressure" as you drink the beverage.

A box has the dimensions of 50 cm × 30 cm × 15 cm

Let each face be A, B and C

The weight of the box = 150 N

Formula for pressure is

P = F/A

To find out which face of the box can withstand a pressure of 0.25 N/cm

we need find the area of each face and find its pleasure

Face A = 50 cm × 30 cm

Area A = l × b

            = 50 × 30

            = 1500 cm²

Pressure A = 150/1500

                  = 0.1 N/cm³

                    0.25 > 0.1

The surface can definitively withstand the pressure of Face A

Face B = 30 cm × 15 cm

Area A = l × b

            = 30 × 15

            = 450 cm²

Pressure A = 150/450

                  = 0.3 N/cm³

                    0.25 < 0.3

The surface could not withstand the pressure of Face B

Face C = 50 cm × 15 cm

Area A = l × b

            = 50 × 15

            = 750 cm²

Pressure A = 150/750

                  = 0.2 N/cm³

                    0.25 > 0.2

The surface can definitively withstand the pressure of Face C

Thus, The surface can only withstand a pressure Face A (50cm × 30cm) and Face C (50cm × 15cm) as the surface can only withstand a pressure of 0.25 N/cm³.

Learn more about Pressure

brainly.com/question/945436

#SPJ9

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