Mail mustafa ba C x Md Sent Mail malubailar x M c Please Show Work And M Mathway

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Mail mustafa ba C x Md Sent Mail malubailar x M c Please Show Work And M Mathway 1Math Probler x Quiz Chap 05 myCourses Blackboard x C www.webassign.net/web/Student/Assignment Responses/submit?dep 15410251 3. 015.5 points I Previous Answers Tiplert 5.PO59 A 2.0 kg block sits on a 4.0 kg block that is on a frictionless table. The coefficients of friction between the blocks are 0.40 and uk 0.10 (a) What is the maximum force F that can be applied to the 4.0 kg block if the 2.0 kg block is not to slide? N 7.84 (b) If F is half this value, find the acceleration of each block. m/s2 (2.0 kg block) m/s 4.0 kg block) Find the magnitude of the force of friction acting on each block. N(2.0 kg block) N (4.0 kg block) (c) If F is twice the value found in (a), find the acceleration of each block. m/s2 (2 kg block) m/s2 (4 kg block)

Explanation / Answer

m2 = 2kg
m4 = 4kg.

Since there is no relative motion between m2 and m4 we use s =0.4.

There are two frictional forces one due to the normal force m2g = m2g
And another due to the normal force m4g = m4g.
The applied force must be equal and opposite to the sum of these frictinal forces.
--------------------------------------...
a)
F = (m2 + m4) g = 0.4*6*9.8 = 23.52 N
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b)
a = force / mass = (0.5*23.52) / 6 = 1.96 m/s^2

The frictionla forces are action-reaction pairs and the frictional force is the same on the two blocks and is given by f = m1 a = 2*1.96 m/s^2 = 3.92 N
--------------------------------------...

c)
If the force is 2*23.52 = 47.04 N, then m2 slips on m4 and the frcional force given by

m2a2 = k*m2g = 0.1*2*9.8 = 1.96 and a2 = 0.1*9.8 = 0.98 m/s^2

Net force = (47.04 -1.96)

Acceleration of 4 kg mass =(47.04 -1.96) / 4 =11.27 m/s^2

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