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3 years ago

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if a load of 1000kg can just be dragged up an incline at 10 degrees to the horizontal by a force of 5KN applied in the most effe

ctive direction,what is the value offthe coefficient offriction?

1 answer:

UkoKoshka [18]3 years ago

5 0

Answer:

The coefficient of friction is 0.34

Explanation:

It is given that,

Mass of the load, m = 1000 kg

It is dragged up an incline at 10 degrees to the horizontal by a force of 5 KN applied in the most effective direction, F = 5 × 10³ N

We need to find the coefficient of friction between the surface and the load. From the attached figure, the load is dragged up with a force of F. A frictional force f will also act in this scenario.

So, $F=f+mg\ sin\theta$

Since, $f=\mu N$

or $f=\mu mg\ cos\theta$

$F=\mu mg\ cos\theta+mg\ sin\theta$

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

$5\times 10^3\ N-1000\ kg\times 9.8\ m/s^2\ sin(10)=\mu mg\ cos\theta$

$\mu=\dfrac{3298.24}{1000\ kg\times 9.8\ m/s^2\times cos(10)}$

$\mu=0.34$

So, the coefficient of friction is 0.34. Hence, this is the required solution.

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

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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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To not exceed the 5,000 characters, see each answer with its explanation below.

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<u><em>Question 1.</em></u>

<em>A feather and a rock dropped at the same time from the same height would land at the same time when dropped by an astronaut on the moon</em>.

The shape and mass of the <em>feather</em> that falls vertically down, through the air, cause the resistance of the air, that opposes the gravitational attraction of the Earth, to be considerably large, compared to the case of the <em>rock</em>.

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<em><u>Question 2.</u></em>

<em>When the soccer ball is kicked, the action and reaction forces do not cancel each other out because the forces act on two different objects.</em>

The third law of Newton, also known as the law of action and reaction, states that whenever an object experience a force (action force) from a second object, the first object will exert a force of equal size and opposite in direction (reaction force) over the second object.

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<u><em>Question 5</em></u>

<em>The bowling ball will have more momentum because it has more mass.</em>

Momentum, P, is the product of the mass (m) and the velocity (v).

$\vec{P}=m\times \vec{v}$

It is clear that the mass of the bowling ball is greater than the mass of the golf ball.

Hence, since they are moving at the same velocity, the product of the mass of the bowling ball times the velocity is greater than the product of the golf mass times the same velocity.

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