Assume the three blocks portrayed in the figure below move on a frictionless sur
ID: 2168649 • Letter: A
Question
Assume the three blocks portrayed in the figure below move on a frictionless surface and a force F = 34-N acts as shown on the m = 3.5-kg block.(a) Determine the acceleration given this system.
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Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. m/s2 (to the right)
(b) Determine the tension in the cord connecting the 3.5-kg and the 1.0-kg blocks.
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Your response differs from the correct answer by more than 10%. Double check your calculations. N
(c) Determine the force exerted by the 1.0-kg block on the 2.0-kg block.
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Your response differs from the correct answer by more than 100%. N
Explanation / Answer
(a) since the surface is frictionless, only external force that acts on the system is F. so, acceleration of system=force on the system/mass of the system=48/(1+2+3.5) m/s^2=7.38 m/s^2 the direction of acceleration is same of that of force, i.e towards +ve x(increasing right side). (b)[let T1 be the tension that acts on the box of 1kg(in right direction) & T2 be be the tension that acts on m(towards left) tension in the cord is same everywhere, assuming it's massless, if it was not so, say T1>T2 then the resulting tension would act on the cord and accelerate it, i.e T1-T2=mass of cord*some acceleration since mass of cord=0, T1-T2=0=>T1=T2=T(say)] There will be contact force P working between 1kg & 2kg boxes for box of 2 kg, P(contact force exerted by 1kg box)=2*7.38 N thus P=14.76 N for box of mass 1kg, T-P(contact force exerted by 2kg box on 1kg box)=1*7.38 T=P+7.38 N=22.14 N (c) already determined, P=14.76 N