ot Question 24 A force ef 64Ns applied 16 an cebjeot and the object is do na obs

Question

ot Question 24 A force ef 64Ns applied 16 an cebjeot and the object is do na observed to accelerate wtN an accelération of 3 0 msit triction is so small that it can be ignored, what is the -massi of the object in kilograms? Show your calculation. Question 2-5 An object of mass 39 kg is observed to acceierate with an acceleration of 2.0 m/s If friction is so small that it can be ignored, what is the orts to emst ni aforce appled to the abject in newtons? Show your calculation The main purpose of RTP Labs 1.3, 1.4, and 1.5 has been to explore the relationship between the forces on an object, its mass, and its acceleration. You have been developing N2L and NiL for one-dimensional situations in which 2e6m tar toomment: direction along the same line Activity 2-3: Newton's Laws in Your Own Words Question 2-6: Express Newton's First Law (the one about constant velocity) in terms of the cambined (net) force applied to an object in your own words clearly and precisely Question 2-7: Express Newton's First Law in equations in terms of the acceleration vector, the combined (net) force vector applied to an object and the object's mass. n Wiley&Sons.; Portions of this material may have been modified locally llam M iMargaret E. Wessling, and Armen N. Kocharian, with assistance from Robert Pelka. Läist modified osresra

Explanation / Answer

Question 2-4: According to Newton's 2nd law of motion, F=ma.

Given F=5.4 N, a=3 m/s2.

Hence m=F/a=5.4/3=1.8 kg.

Question 2-5: According to Newton's 2nd law of motion, F=ma.

Given m=39kg, a=2 m/s2.

Hence Force applied is F=39*2=78 N.

Question 2-6: An object is said to be in it's state of rest or of uniform motion unless there is a change in the net force acting on the object.

Question 2-7: F=ma, a=F/m.

Acceleration a=dv/dt. Hence dv=a*dt.

Integrating on both sides, we get v=at+constant(integration constant)

Here a=F/m. Hence v=F/m+constant

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