..000 cricket 8:28 AM session.masteringphysics.com C Exercise 4.60 13 of 14 C Ex
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..000 cricket 8:28 AM session.masteringphysics.com C Exercise 4.60 13 of 14 C Exercise 4.60 Unless othenwise stated, all objects are located near the Earh's surface, where g = 980 m/s In the frictionless apparatus shown in the figure 0 37 Q Tap image to zoom m1=20 kg Part A What is both masses are at rest Express your answer using Ewo significant figures. kg Submit Give Up Part B How about if both masses are moving at constant velocity! Express your answer using two significant figures. kg Submit iveExplanation / Answer
Weight of first block m1 will act vertically downwards. Then dividing its weight into components:
1st component: Acting down the slope, parallel to the slope,
W11= m1gsin37o= (2.0)(9.8)sin37o= 11.79N...................................(1)
2nd component: Acting perpendicular to the slope in downward direction,
W12= m1gcos37o= (2.0)(9.8)cos37o= 15.65N
Weight of the 2nd block, W2= m2g
Let Tension in the string be "T", then using Balance of Forces concept for 2nd block,
T= W2= m2g (Since the block is at rest)........................(2)
Similarly using the balance of forces concept for 1st block along the slope,
T= W11= 11.79 (Using equation 1)
using equation 2 in above equation, (since string is same hence Tension in the string will be same)
m2g= 11.79
m2= 11.79/9.80= 1.203kg (ANS)
b). Now if both blocks are moving at constant speed that means that there net acceleration is 0. Hence Net force acting on the system will be again 0. Thus we'll get the same situation just like that in part (a) but this time moving with constant velocity and thereby with 0 acceleration.
Hence mass of 2nd block will be same, i.e.
m2= 1.203kg (ANS)
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