TPR SW Physics #179

[crazy person]

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Hey guys,

I'm a little bit confused on this problem. If the block started from rest at the top of the incline and its asking with what speed does it reach the bottom? Then PE is transferred to KE at the bottom, then it should be v = square root 2gh to find the speed at the bottom. Why, does the answer use work energy theorem which gives a different answer?

Thanks !!!

 

[dudewheresmymd]

Both methods give the same answer (as it should as long as the laws of physics are not being violated 🙂)


Method 1: mgh=1/2mv^2 --> sqrt (2gh) = final velocity
sqrt (2gh) = sqrt( 2*10*height of the ramp =1.5m)

30,60,90 triangle --> sin 30 = opposite height of ramp /hypotenuse (3m)

(.5)(3)= height of ramp = 1.5 m

sqrt (2*10*1.5)= sqrt (30) aprox = 5.5 m/s

Or you can do their way, since the work that gravity does = force * displacement therefore,

work= mgsintehta *d = 2*10*sin(30) * 3 = 30 J OR just do change in mgh since gravity is a conservative force, delta pe = mgh = 2*10* (1.5m height of ramp) = 30J

once you know the work, they set it equal to the change in the kinetic energy, also valid, as there are no external forces acting on the system,

Wnet= delta ke
30 J= 1/2m(vf^2-vi^2); vi^2=0 (from rest), multiply both sides by 2
60=2kg*(vf^2)
30 = vf^2
sqrt (30)=vf=5.5 m/s

sometimes there is more than one to do these energy questions, pick the most convenient way for you and be confident that your way is as equally valid 🙂

I would do the first method as it saves time and involves less calculation for this.

 

[crazy person]

THANK YOU !!! you are right, the first method saves more time !🙂