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Andreas93 [3]
3 years ago
14

You are standing a distance of 17.0 meters from the center of a merry-go-round. The merry-go-round takes 9.50 seconds to go comp

letely around once and you have a mass of 55.0 kg.
a. What will be your speed as you move around the center of the merry-go-round?
b. What will be your centripetal acceleration as you move around the center of the merry-go-round?
c. What will be the magnitude of the centripetal force necessary to keep your body moving around the center of this merry-go-round at the calculated speed?
d. How much frictional force will be applied to you by the surface of the merry-go-round?
e. what is the minimum coefficient of friction between your shoes and the surface of the merry-go-round?
Physics
1 answer:
GenaCL600 [577]3 years ago
3 0

Answer:

(a)11.24 m/s

(b)7.44 m/s

(c)409 N

(d)539.55\mu

(e) 0

Explanation:

The period for 1 circle 2\pi of the merry go around is 9.5s. It means the angular speed is:

\omega = \theta / t = 2\pi / 9.5 \approx 0.661 rad/s

(a)The speed is

v = \omega * R = 0.661 * 17 = 11.24 m/s

(b) Centripetal acceleration:

a = \frac{v^2}{R} = \frac{11.24^2}{17} = 7.44 m/s^2

(c) Magnitude of the force that keeps you go around at this acceleration

F = ma = 55 * 7.44 = 409 N

(d) let the coefficient of friction by \mu. The frictional force shall be this coefficient multiplied by normal force reverting gravity of the man

F_f = mg\mu = 55*9.81\mu = 539.55\mu

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Answer:

The acceleration of the crate is 1.8 m/s² so the answer is a.

Explanation:

The very first thing you must do when solving this problem is to draw a free body diagram. (The body diagram is attached to this answer)

So once we got the free body diagram, we can analyze it and build our sum of forces in the x and y directions. Notice that according to the diagram, there are 4 forces to this problem, Normal (N), Weight (W), kinetic friction (fk) and the 750N force.

As one may see in the free body diagram, two of the forces are vertical forces: N and W, so we can use them to build a sum of forces:

Starting with the sum of forces in the y-direction, we get:

ΣF_{y}=0

We set the sum equal to zero because there is no movement in the y-direction, so the system is in vertical equilibrium.

so the sum will be:

N-W=0

when solving for N we get that:

N=W

where W is found by multiplying the mass of the crate by the acceleration of gravity:

N=250kg*9.8m/s²

N=2450N

Once we found the normal force, we can use it to find the kinetic friction which is given by the following formula:

f_{k}=Nμ

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f_{k}=2450N*0.12

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f_{k}=294

With this information we can go ahead and find the sum of horizontal forces:

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In this case the sum is equal to mass times acceleration because the crate is moving horizontally due to the action of a force, so it will have an acceleration.

so the sum of forces look like this:

750N-f_{k}=ma

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when solving for a we get:

a=\frac{759N-294N}{250kg}\\ \\a=1.8m/s^{2}

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