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3 years ago

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A 150kg motorcycle starts from rest and accelerates at a constant rate along with a distance of 350m. The applied force is 250N

and the coefficient of kinetic friction is 0.03. What is the net force applied to the motorcycle?

1 answer:

mart [117]3 years ago

3 0

Answer:

205N

Explanation:

The net force (F) is the difference between the applied force( $F_{A}$ ) and the kinetic frictional force( $F_{R}$ ). i.e

F = $F_{A}$ - $F_{R}$ -----------------(i)

Note that;

$F_{R}$ = μmg

Where;

μ = coefficient of kinetic friction

m = mass of the body

g = acceleration due to gravity = 10m/s²

Equation (i) then becomes;

F = $F_{A}$ - μmg -------------------(ii)

<em>Given from question;</em>

m = mass of motorcycle = 150kg

μ = 0.03

$F_{A}$ = 250N

Substitute these values into equation (ii) as follows;

F = 250 - (0.03 x 150 x 10)

F = 250 - (45)

F = 205N

Therefore, the net force applied to the motorcycle is 205N

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3 years ago

a ball of mass 100g moving at a velocity of 100m/s collides with another ball of mass 400g moving at 50m/s in same direction, if

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Velocity of the two balls after collision: $60\; \rm m \cdot s^{-1}$ .

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Explanation:

<h3>Velocity</h3>

Because the question asked about energy, convert all units to standard units to keep the calculation simple:

Mass of the first ball: $100\; \rm g = 0.1\; \rm kg$ .
Mass of the second ball: $400\; \rm g = 0.4 \; \rm kg$ .

The two balls stick to each other after the collision. In other words, this collision is a perfectly inelastic collision. Kinetic energy will not be conserved. The velocity of the two balls after the collision can only be found using the conservation of momentum.

Assume that the system of the two balls is isolated. Thus, the sum of the momentum of the two balls will stay the same before and after the collision.

The momentum of an object of mass $m$ and velocity $v$ is: $p = m \cdot v$ .

Momentum of the two balls before collision:

First ball: $p = m \cdot v = 0.1\; \rm kg \times 100\; \rm m \cdot s^{-1} = 10\; \rm kg \cdot m \cdot s^{-1}$ .
Second ball: $p = m \cdot v = 0.4\; \rm kg \times 50\; \rm m \cdot s^{-1} = 20\; \rm kg \cdot m \cdot s^{-1}$ .
Sum: $10\; \rm kg \cdot m \cdot s^{-1} + 20 \; \rm kg \cdot m \cdot s^{-1} = 30 \; \rm kg \cdot m \cdot s^{-1}$ given that the two balls are moving in the same direction.

Based on the assumptions, the sum of the momentum of the two balls after collision should also be $30\; \rm kg \cdot m \cdot s^{-1}$ . The mass of the two balls, combined, is $0.1\; \rm kg + 0.4\; \rm kg = 0.5\; \rm kg$ . Let the velocity of the two balls after the collision $v\; \rm m \cdot s^{-1}$ . (There's only one velocity because the collision had sticked the two balls to each other.)

Momentum after the collision from $p = m \cdot v$ : $(0.5\, v)\; \rm kg \cdot m \cdot s^{-1$ .
Momentum after the collision from the conservation of momentum: $30\; \rm kg \cdot m \cdot s^{-1}$ .

These two values are supposed to describe the same quantity: the sum of the momentum of the two balls after the collision. They should be equal to each other. That gives the equation about $v$ :

$0.5\, v = 30$ .

$v = 60$ .

In other words, the velocity of the two balls right after the collision should be $60\; \rm m \cdot s^{-1}$ .

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Sum of the kinetic energies of the two balls right after the collision:

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Answer:

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3 years ago

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Answer:

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Explanation:

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