All categories

3 years ago

In order to get the bike to move and stay moving, what friction must be overcome

1 answer:

4 0

Friction is the resistance offered against the motion of one surface over the other. This is because, at the microscopic level, each surface has hills and valleys i.e. the surfaces are not completely smooth and hence the roughness of one surface locks over the other and restricts its motion. There are two kinds of friction: static friction and kinetic friction. When the object is at rest, static friction acts. While the object is in motion, kinetic friction acts. Static friction is greater than the kinetic friction because a larger force is required to change the state of rest to motion. You must have experienced that if you do not accelerate the bike while its in motion, it would slow down and eventually stop. In order to get the bike moving, initially static friction must be overcome and thereafter, kinetic friction to make it stay in motion.

You might be interested in

A fireman is sliding down a fire pole. As he speeds up, he tightens his grip on the pole, thus increasing the vertical frictiona

Andrei [34K]

Answer:

The fireman will continue to descend, but with a constant speed.

Explanation:

In kinetic friction (which is the case discussed here) since the fireman is already in motion because of a certain force, once the frictional force matches the normal force, the fireman will stop accelerating and continue moving at a constant rate with the original speed he had. We will need a force greater than the normal force acting on the fireman to cause a deceleration.

We need to understand the difference between static friction and kinetic friction.

Static friction occurs in objects that are stationary, while kinetic friction occurs in objects that are already in motion.

In static friction, when the frictional force matches the weight or normal force of the object, the object remains stationary.

While in kinetic friction, when the frictional force matches the normal force, the object will stop accelerating. This is the case of the fireman sliding down the pole as discussed above.

8 0

3 years ago

A 2.00 kg block on a horizontal floor is attached to a horizontal spring that is initially compressed 0.0300 m . The spring has

iogann1982 [59]

Answer:

v = 0.41 m/s

Explanation:

In this case, the change in the mechanical energy, is equal to the work done by the fricition force on the block.
At any point, the total mechanical energy is the sum of the kinetic energy plus the elastic potential energy.
So, we can write the following general equation, taking the initial and final values of the energies:

$\Delta K + \Delta U = W_{ffr} (1)$

Since the block and spring start at rest, the change in the kinetic energy is just the final kinetic energy value, Kf.
⇒ Kf = 1/2*m*vf² (2)
The change in the potential energy, can be written as follows:

$\Delta U = U_{f} - U_{o} = \frac{1}{2} * k * (x_{f} ^{2} - x_{0} ^{2} ) (3)$

where k = force constant = 815 N/m

xf = final displacement of the block = 0.01 m (taking as x=0 the position

for the spring at equilibrium)

x₀ = initial displacement of the block = 0.03 m

Regarding the work done by the force of friction, it can be written as follows:

$W_{ffr} = - \mu_{k}* F_{n} * \Delta x (4)$

where μk = coefficient of kinettic friction, Fn = normal force, and Δx =

horizontal displacement.

Since the surface is horizontal, and no acceleration is present in the vertical direction, the normal force must be equal and opposite to the force due to gravity, Fg:
Fn = Fg= m*g (5)
Replacing (5) in (4), and (3) and (4) in (1), and rearranging, we get:

$\frac{1}{2} * m* v^{2} = W_{ffr} - \Delta U = W_{ffr} - (U_{f} -U_{o}) (6)$

$\frac{1}{2} * m* v^{2} = (- \mu_{k}* m*g* \Delta x) -\frac{1}{2} * k * (x_{f} ^{2} - x_{0} ^{2} ) (7)$

Replacing by the values of m, k, g, xf and x₀, in (7) and solving for v, we finally get:

$\frac{1}{2} * 2.00 kg* v^{2} = (-0.4*2.00 kg*9.8m/s2*0.02m) +( (\frac{1}{2} *815 N/m)* (0.03m)^{2} - (0.01m)^{2}) = -0.1568 J + 0.326 J (8)$

$v =\sqrt{(0.326-0.1568} = 0.41 m/s (9)$

7 0

3 years ago

In what way are gravitational and electrical forces similar?

Nadusha1986 [10]

Answer:

D. Both occur between objects independently whether they are in contact or not.

Explanation:

- The gravitational force is a force that is exerted between two (or more) objects having mass. This force is always attractive and its magnitude is given by

$F=G\frac{m_1 m_2}{r^2}$

where G is the gravitational constant, m1 and m2 are the two masses, and r is the distance between the two masses.

- The electrical force is a force that is exerted between two (or more) objects having electrical charge. It can be either attractive or repulsive, depending on the sign of the two charges, and its magnitude is given by

$F=k\frac{q_1 q_2}{r^2}$

where k is the Coulomb's constant, q1 and q2 are the two charges, and r the distance between the two charges.

Looking at both formulas, we see that the two forces are present even when the two objects are not in contact with each other (in fact, r can assume any value in the formula). They are said to be non-contact forces. Therefore, the correct option is

D. Both occur between objects independently whether they are in contact or not.

6 0

3 years ago

Air in the atmosphere is heated by the ground this warm air then rises and cooler air falls this is an example of what type of p

Alona [7]

The answer for both is ‘B’

8 0

3 years ago

Fossils are dated using radioactive... a. hydrogen b. helium c. carbon

OlgaM077 [116]

Answer:

carbon

Explanation:

7 0

3 years ago

Read 2 more answers