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3 years ago

8

What is the magnitude of the force needed to keep a 60 newton rubber block moving across level,dry asphalt in a straight line at

a constant speed of 2.0 meters per second

2 answers:

pantera1 [17]3 years ago

6 0

let Coefficients of Friction of Rubber on asphalt (dry) =0.7

F= Coefficients of Friction * normal force = 0.7 * 60 =42 N

so the net force of the rubber is zero, meaning it will travel at a constant speed.

When the rubber is travel at 2m/s, 42N is required to keep moving at constant speed

MissTica3 years ago

5 0

Answer:

F = 40.2 N

Explanation:

Weight of the rubber block, W = 60 N

If m is the mass of the rubber block,

$m=\dfrac{W}{g}$

$m=\dfrac{60\ N}{9.8\ m/s^2}$

m = 6.12 kg

We know that the coefficient of friction between rubber block and dry asphalt is, $\mu_k=0.67$

The relation between the coefficient of friction and the force is given by :

$F=\mu_k N$

Here, N = W

$F=\mu_k W$

$F=0.67\times 60\ N$

F = 40.2 N

So, the magnitude of force needed to keep a 60 newton rubber block moving across level,dry asphalt in a straight line is 40.2 N. Hence, this is the required solution.

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A 0.405 kg mass is attached to a spring with a force constant of 26.3 N/m and released from rest a distance of 3.31 cm from the

miv72 [106K]

Answer:

0.231 m/s

Explanation:

m = mass attached to the spring = 0.405 kg

k = spring constant of spring = 26.3 N/m

x₀ = initial position = 3.31 cm = 0.0331 m

x = final position = (0.5) x₀ = (0.5) (0.0331) = 0.01655 m

v₀ = initial speed = 0 m/s

v = final speed = ?

Using conservation of energy

Initial kinetic energy + initial spring energy = Final kinetic energy + final spring energy

(0.5) m v₀² + (0.5) k x₀² = (0.5) m v² + (0.5) k x²

m v₀² + k x₀² = m v² + k x²

(0.405) (0)² + (26.3) (0.0331)² = (0.405) v² + (26.3) (0.01655)²

v = 0.231 m/s

8 0

3 years ago

An object with velocity 141 ft/s has a kinetic energy of 1558.71 ftâˆ™lbf, on a planet whose gravity is 31.5 ft/s2. What is its

Sidana [21]

Answer:

The mass of the object is 5.045 lbm.

Explanation:

Given;

kinetic energy of the object, K.E = 1558.71 ft.lbf

velocity of the object, V = 141 ft/s

The kinetic energy of the object is calculated as;

$K.E = \frac{1}{2} mV^2\\\\mV^2 = 2K.E\\\\m = \frac{2K.E}{V^2} \\\\1 \ lbf = 32.174 \ lbm.ft/s^2\\\\m = \frac{2 \ \times \ 1558.71 \ ft.lbf \ \times \ 32.174 \ lbm.ft/s^2 }{(141 \ ft/s)2 \ \ \times \ \ \ \ 1 \ lbf\ }$

$m = \frac{(2 \ \times \ 1558.71 \ \times \ 32.174) \ lbm.ft^2/s^2 }{(141 )^2\ ft^2/s^2 }\\\\m = \frac{(2 \ \times \ 1558.71 \ \times \ 32.174) \ lbm }{(141 )^2 }\\\\m = 5.045 \ lbm$

Therefore, the mass of the object is 5.045 lbm.

6 0

2 years ago

While entering a freeway, a car accelerates from rest at a rate of 2.40 m/s2 for 12.0 s. (a) Draw a sketch of the situation. (b)

ArbitrLikvidat [17]

Answer:

a) See attached picture, b) We know the initial velocity = 0, initial position=0, time=12.0s, acceleration= $2.40m/s^{2}$ , c) the car travels 172.8m in those 12 seconds, d) The car's final velocity is 28.8m/s

Explanation:

a) In order to draw a sketch of the situation, I must include the data I know, the data I would like to know and a drawing of the car including the direction of the movement and its acceleration, just like in the attached picture.

b) From the information given by the problem I know:

initial velocity =0

acceleration = $2.40m/s^{2}$

time = 12.0 s

initial position = 0

c)

unknown:

displacement.

in order to choose the appropriate equation, I must take the knowns and the unknown and look for a formula I can use to solve for the unknown. I know the initial velocity, initial position, time, acceleration and I want to find out the displacement. The formula that contains all this data is the following:

$x=x_{0}+V_{x0}t+\frac{1}{2}a_{x}t^{2}$

Once I got the equation I need to find the displacement, I can plug the known values in, like this:

$x=0+0(12s)+\frac{1}{2}(2.40\frac{m}{s^{2}} )(12s)^{2}$

after cancelling the pertinent units, I get that my answer will be given in meters. So I get:

$x=\frac{1}{2} (2.40\frac{m}{s^{2}} )(12s)^{2}$

which solves to:

$x=172.8m$

So the displacement of the car in 12 seconds is 172.8m, which makes sense taking into account that it will be accelerating for 12 seconds and each second its velocity will increase by 2.4m/s.

d) So, like the previous part of the problem, I know the initial position of the car, the time it travels, the initial velocity and its acceleration. Now I also know what its final position is, so we have more than enough information to find this answer out.

I need to find the final velocity, so I need to use an equation that will use some or all of the known data and the unknown. In order to solve this problem, I can use the following equation:

$a=\frac{V_{f}-V_{0} }{t}$

Next, since I need to find the final velocity, I can solve the equation just for that, I can start by multiplying both sides by t so I get:

$at=V_{f}-V_{0}$

and finally I can add $V_{0}$ to both sides so I get:

$V_{f}=at+V_{0}$

and now I can proceed and substitute the known values:

$V_{f}=at+V_{0}$

$V_{f}=(2.40\frac{m}{s^{2}}} (12s)+0$

which solves to:

$V_{f}=28.8m/s$

8 0

3 years ago

Read 2 more answers

I drop an egg from a certain distance and it takes the egg 3.74 seconds to reach the ground. How high up was the egg?

Ulleksa [173]

Answer:

<em>B. 68.6m</em>

Explanation:

<u>Free Fall Motion </u>

When a body is left to move in the air with no friction, the motion is ruled only by the force of gravity. The vertical distance a body travels in the air after a time t is .

$\displaystyle y=\frac{gt^2}{2}$

We know the egg takes 3.74 seconds to reach the ground. The height it was launched from is

$\displaystyle y=\frac{(9.8)(3.474)^2}{2}$

$\displaystyle y=68.54\ m$

The closest correct option is

B. 68.6m

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3 years ago

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kow [346]

Answer:

Conservation of angular momentum

Explanation:

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3 years ago