You throw a 0.2 kg ball at a 47degree angle with a speed of 35 m/s. Using an ene
ID: 1281073 • Letter: Y
Question
You throw a 0.2 kg ball at a 47degree angle with a speed of 35 m/s. Using an energy analysis find the maximum height the ball reaches relative to the release point. Don't use the kinematic equations, they're harder to use anyway. A cart of mass M = 3.5 kg is released from rest on a track that makes an angle 0 = 32Degree with the horizontal. The cart is connected to a hanging mass m=2.5 kg by a string wrapped over a pulley. The wheels are quite sticky and have effective coefficient of kinetic friction of 0.2. The cart travels a distance of 0.7 m down the track. Directly calculate the work done on M by all forces that act on M. By directly calculate I mean use the integral definition Having calculated all the individual works done by all the individual forces, verify that the work-kinetic energy theorem holds true.Explanation / Answer
5)The velocity at maximum height=ux0=35Cos47=23.87m/s
Conserving energy between highest and lowest points.
0.5mu^2=mgh+0.5mux02
0.5*35^2=9.81*h+0.5*23.87^2
h=33.4m
1)Friction on M=uMgCos(theta)=0.2*3.5*9.81*Cos32=5.824N
For m,
mg-T=ma
2.5*9.8-T=2.5a
For M,
MgSin(theta)-uMgCos(theta)+T=Ma
3.5*9.81*Sin32-0.2*3.5*Cos32=3.5a
T=9.137N, a=6.145
Work by friction=-5.824*0.7=-4.08J
Work by Tension on M=9.137*0.7=6.40J
Work by tension on m=-9.137*0.7=-6.4J
Work by gravity on m=2.5*0.7*9.81=17.17J
Work by gravity on M=3.5*9.81*0.7Sin32=12.74N
Total work done on system=-4.08J+17.17J+12.74J=25.83 J
Final velocity=sqrt(2as)=sqrt(2*6.145*0.7)=2.933m/s
Kinetic energy=0.5*(2.5+3.5)*2,933^2=25.81J
Since all the work done on the blocks in converted into kinetic energy, the work- KE theorem holds.
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