Jane and Jon, with masses of 50 kg and 60kg, respectively,stand on a frictionles

Question

Jane and Jon, with masses of 50 kg and 60kg, respectively,stand on a frictionless surface 10m apart. John pulls on a ropethat connects fim to jane, giving Jane an acceleration of 0.92m/s2 toward him.    a) What is JOhn's acceleration?    b) If the pulling force is appliedconstantly, where will Jane and JOhn meet? Jane and Jon, with masses of 50 kg and 60kg, respectively,stand on a frictionless surface 10m apart. John pulls on a ropethat connects fim to jane, giving Jane an acceleration of 0.92m/s2 toward him.    a) What is JOhn's acceleration?    b) If the pulling force is appliedconstantly, where will Jane and JOhn meet?

Explanation / Answer

a) F=m*a      F=50kg*.92m/s^2    F=46N    46N=60kg*a    46N/60kg=a    a=.767m/s^s b) .92m/s^2 + .767m/s^2 = 1.687m/s^2    s=.5*a*t^2    10m=.5*1.687m/s^2*t^2    10m/(.5*1.687m/s^2)=t^2    (10m/(.5*1.687m/s^2))^.5=t    t=3.44s s=.5*.92m/s^2*3.44s s=5.45m or s=.5*.767m/s^2*3.44s a=4.55m So, they will meet 5.45m from Jane's intial position, or 4.55mfrom John's initial postion. Hope that helped. b) .92m/s^2 + .767m/s^2 = 1.687m/s^2    s=.5*a*t^2    10m=.5*1.687m/s^2*t^2    10m/(.5*1.687m/s^2)=t^2    (10m/(.5*1.687m/s^2))^.5=t    t=3.44s s=.5*.92m/s^2*3.44s s=5.45m or s=.5*.767m/s^2*3.44s a=4.55m So, they will meet 5.45m from Jane's intial position, or 4.55mfrom John's initial postion. Hope that helped.

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