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A certain light truck can go around a flat curve having a radius of 150 m with a maximum speed of 35.5 m/s. a) What is the coeff

postnew [5]

Answer:

The coefficient of friction present between the roadway and the wheels of the truck is 0.833.

Explanation:

Given:

Radius of the curve (R) = 150 m

Maximum speed of truck (v) = 35.5 m/s

Let the coefficient of friction between the roadway and the wheels of the truck be "μ".

As the truck is moving around a circular curve. So, the force acting on it is centripetal force which acts in the radial inward direction towards the center of the circular curve.

The centripetal force acting on the truck is given as:

$F_c=\frac{mv^2}{R}$

Now, the friction between the roadway and the wheels of the truck is responsible for providing the necessary centripetal force. So, frictional force is equal to the centripetal force necessary for circular motion.

Frictional force is given as:

$f=\mu N$

Where, 'N' is the normal force. Since there is no vertical motion, the normal force is equal to weight of truck. So,

$N=mg$

Therefore, frictional force, $f=\mu mg$

Now, frictional force = centripetal force

$f=F_c\\\\\mu mg=\frac{mv^2}{R}\\\\\mu = \frac{v^2}{Rg}$

Plug in the given values and solve for 'μ'. This gives,

$\mu=\frac{(35\ m/s)^2}{(150\ m)(9.8\ m/s^2)}\\\\\mu=\frac{1225\ m^2/s^2}{1470\ m^2/s^2}\\\\\mu=0.833$

Therefore, the coefficient of friction present between the roadway and the wheels of the truck is 0.833

7 0

3 years ago

A 60 kg runner accelerates during a race at 3m/s2. what force is exerted on the earth with each step

marusya05 [52]

Answer:

180 Newton(N)

Explanation:

force =mass *acceleration

=60 * 3

=180 kgm/s^2

=180 N

6 0

3 years ago

1. A student lifts a box of books that weighs 185 N. The box is

aksik [14]

1) 148 J

When lifting an object, the work done on the object is equal to its change in gravitational potential energy. Mathematically:

$W = \Delta U = (mg) \Delta h$

where

mg is the weight of the object

$\Delta h$ is the change in height

For the box in this problem,

mg = 185 N

$\Delta h = 0.800 m$

Substituting into the equation, we find:

$W=(185)(0.800)=148 J$

2) (a) 28875 J

The work done by a force applied parallel to the direction of motion of the object is given by

$W=Fd$

where

F is the magnitude of the force

d is the displacement

In this problem,

F = 825 N is the force applied by the two students together

d = 35 m is the displacement of the car

Substituting,

$W=(825)(35)=28875 J$

2) (b) 57750 J

As seen previously, the equation that gives the work done by the force is

$W=Fd$

We see that the work done is proportional to the magnitude of the force: therefore, if the force is doubled, then the work done is also doubled.

The work done previously was

W = 28875 J

Now the force is doubled, so the new work done will be

$W' = 2(28875)=57750 J$

3) 4.4 J

In this case, the force acting on the ball is the force of gravity, whose magnitude is:

$F = mg$

where

m = 0.180 kg is the mass of the ball

$g=9.8 m/s^2$ is the acceleration of gravity

Solving the equation,

$F=(0.180)(9.8)=1.76 N$

Now we find the work done by gravity using the same formula applied before:

$W=Fd$

where d = 2.5 m is the displacement of the ball. We can apply this version of the formula since the force is parallel to the displacement. Substituting,

$W=(1.76)(2.5)=4.4 J$

4) 595.2 kg

In this case, we have the work done on the box:

W = 7.0 kJ = 7000 J

And we also know the change in height of the box:

$\Delta h = 1.2 m$

As we stated in part a), the work done on the box is equal to its change in gravitational potential energy:

$W=mg \Delta h$

Solving for m, we find

$m=\frac{W}{g \Delta h}$

And substituting the numerical values, we find the mass of the box:

$m=\frac{7000}{(9.8)(1.2)}=595.2 kg$

5) They do the same work

In fact, the net work done by each person on the box is equal to the change in gravitational potential energy of the box:

$W=mg \Delta h$

Where $\Delta h$ is the difference in height between the final position and the initial position of the box.

This means that the work done on the box depends only on its initial and final position, not on the path taken. The two men carry the box along different paths, however the reach at the end the same position, and they started from the same position: this means that the value of $\Delta h$ is the same for both of them, so the work they have done is exactly the same.

5 0

3 years ago

A bicyclist in the Tour de France crests a mountain pass as he moves at 18 km/h. At the bottom, 4.0 km farther, his speed is at

Allisa [31]

We are given:

v0 = initial velocity = 18 km/h

d = distance = 4 km

v = final velocity = 75 km/h

a =?

We can solve this problem by using the formula:

v^2 = v0^2 + 2 a d

75^2 = 18^2 + 2 (a) * 4

5625 = 324 + 8a

a = 662.625 km/h^2

6 0

3 years ago

Suppose the displacement of an object is related to time according to the expression x=By*2, what are the dimensions of B

laiz [17]

Answer:

3

Explanation:

AP3X

5 0

3 years ago