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fiasKO [112]
3 years ago
12

Calculate the maximal friction force for a parked car between the rubber tires and a wet street. Assume the car’s mass is 1600 k

g with a weight of 15680 N. The static friction coefficient for rubber and wet asphalt is μs=0.8.
Physics
1 answer:
NikAS [45]3 years ago
5 0

Answer:

Fr=12544N

Explanation:

1. Find the equation of eht maximal friction force:

The maximal friction force is given by the equation Fr=usmg, where μs is the static friction coefficient, m is the car´s mass and g is the gravitational force.

2. Replace values in the equation to find the answer:

Fr=0.8*1600kg*9.8\frac{m}{s^{2}}

Fr=12544N

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Suppose the ski patrol lowers a rescue sled and victim, having a total mass of 90.0 kg, down a p slope at constant speed, as sho
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The work done by friction to move the sled is  - 1,323 J.

<h3>What is Coefficient of friction?</h3>
  • The friction coefficient is the ratio of the normal force pressing two surfaces together to the frictional force preventing motion between them.
  • Typically, it is represented by the Greek letter µ. In terms of math, is equal to F/N, where F stands for frictional force and N for normal force.
  • The coefficient of friction has no dimensions because both F and N are measured in units of force (such as newtons or pounds). For both static and kinetic friction, the coefficient of friction has a range of values.
  • When an object experiences static friction, the frictional force resists any applied force, causing the object to stay at rest until the static frictional force is removed. The frictional force opposes an object's motion in kinetic friction.

Solution:

Given that

Coefficient of friction (µ) = 0.10

Mass (m) = 90kg

distance covered (d) = 30m

We use the formula:

friction work = -µmgdcos∅

friction work = -0.100 × 90 kg × 9.8 m/s² × 30 m × cos 60°

friction work = - 1,323 J

Know more about Coefficient of friction numerical brainly.com/question/19308401

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A 5.50 kg sled is initially at rest on a frictionless horizontal road. The sled is pulled a distance of 3.20 m by a force of 25.
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(a) 69.3 J

The work done by the applied force is given by:

W=Fd cos \theta

where:

F = 25.0 N is the magnitude of the applied force

d = 3.20 m is the displacement of the sled

\theta=30^{\circ} is the angle between the direction of the force and the displacement of the sled

Substituting numbers into the formula, we find

W=(25.0 N)(3.20 m)(cos 30^{\circ})=69.3 J

(b) 0

The problem says that the surface is frictionless: this means that no friction is acting on the sled, therefore the energy dissipated by friction must be zero.

(c) 69.3 J

According to the work-energy theorem, the work done by the applied force is equal to the change in kinetic energy of the sled:

\Delta K = W

where

\Delta K is the change in kinetic energy

W is the work done

Since we already calculated W in part (a):

W = 69.3 J

We therefore know that the change in kinetic energy of the sled is equal to this value:

\Delta K=69.3 J

(d) 4.9 m/s

The change in kinetic energy of the sled can be rewritten as:

\Delta K=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2 (1)

where

Kf is the final kinetic energy

Ki is the initial kinetic energy

m = 5.50 kg is the mass of the sled

u = 0 is the initial speed of the sled

v = ? is the final speed of the sled

We can calculate the variation of kinetic energy of the sled, \Delta K, after it has travelled for d=3 m. Using the work-energy theorem again, we find

\Delta K= W = Fd cos \theta =(25.0 N)(3.0 m)(cos 30^{\circ})=65.0 J

And substituting into (1) and re-arrangin the equation, we find

v=\sqrt{\frac{2 \Delta K}{m}}=\sqrt{\frac{2(65.0 J)}{5.50 kg}}=4.9 m/s

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