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Homepage Carmen x KHomepage Carmei X Hwy 4 x Home I Chegg.com x C D www.webassign.net/web/Student/Assignment-Responses/last?dep 10894588 Apps T 58 phrases That... Oclubs O Students Buck.. O ENG 1182 We Jeon, Ryan L-... IL Carmen Login Q Physics 1250 ABioChemistry 4. OHigh On Tropic Q Other bookmarks 6. 13 points I Previous Answers My Notes An inventive child named Nick wants to reach an apple in a tree without climbing the tree. Sitting in a chair connected to a rope that passes over a frictionless pulley (see figure below), Nick pulls on the loose end of the rope with such a force that the spring scale reads 270 N. Nick's true weight is 310 N, and the chair weighs 160 N. Nick's feet are not touching the ground. (b) Find Nick's acceleration, using upward as positive. 146 m/s (c) Find the magnitude of the force Nick exerts on the chair. 3.81 Your response differs from the correct answer by more than 100%. N (d") Instead Nick hands the rope with the scale to his friend Barney, who stands on the ground. Barney pulls on the rope so that the spring scale again reads 270 N. What is Nick's acceleration now, again using upward as positive. T. -13 points My Notes Two objects are connected by a light string that passes over a frictionless pulley as shown in the figure below. Assume the incline is frictionless and take m1 2.00 kg, m2 6.05 kg, and 6 54.5°. (b) Find the magnitude of the accelerations of the objects. m/s2 (c) Find the tension in the string (d) Find the speed of each object 3.00 s after being released from rest. m/s :24 PM 2/12/2015

Explanation / Answer

a) We first find the combined weight of Chris and the chair: 310N+160N=470N. Next we solve for m in F=ma. m=F/a=470N/9.8m/s^2=47.95kg. We now solve for a since we are the force that the scale reads. a=F/m=270N/47.95kg=5.63m/s^s.

b) I'm not 100% of this, but here it goes. We first find the mass of the chair. m=F/a=160N/9.8m/s^2=16.32kg. We now find the mass of Chris. m=47.95kg-16.32kg=31.63kg. The force is moving up so the force is directed upward and equal to ma=31.63kg*5.63m/s^2=178.07N. Chris' weight causes a downward force of ma=31.63kg*9.8m/s^2=309.974N. We used a of gravity because Chris body is fighting the chair from moving up. The force the chair feels is the sum of these two=178.07N+309.974N=488.044N.

c.For the last one, the tension of 270is less than the combined weight of the chair plus Nick a=0

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1.

  This is pretty simple, the acceleration of both blocks must be the same (otherwise the string stretches or compresses--i.e. this is the only way to keep the string "taught").

There are technically three forces on mass2: tension, the normal force, and gravity...

...if you remember from previous problems the component of gravity along the incline is given by: mgsin(?). This takes care of the normal force (because it exactly cancels the force of gravity in the direction perpendicular to the incline), leaving only two forces on mass 2:

F? = m?gsin(?) - T = m?a

Looking at mass 1, there are only two forces on mass one: the tension and gravity--both in the same direction (well vertical direction--they oppose each other). Tension is up and gravity is down...since mass 2 is heavier let's suspect that the acceleration is up (although this makes no difference--if we guess wrong, we'll get a negative tension and acceleration):

F? = T - m?g = m?a
--> so this gives two equations which can be used to find the acceleration

T - m?g = m?a
m?gsin(?) - T = m?a
--> add the two equations to cancel T (the tension)

m?gsin(?) - m?g = m?a + m?a
--> factor out a and solve: m?a - m?a = (m? + m?) * a

a = g(m?sin(?) - m?) / (m? + m?)
-->

a = 9.8 * (6.05 * sin(54.5

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