1. The force on a 0.500-kg particle depends on position such that Fr) (1.00)x2 i
ID: 1583488 • Letter: 1
Question
1. The force on a 0.500-kg particle depends on position such that Fr) (1.00)x2 i (4.00)y j N. If (0,0), what will be its speed when it reaches the position r = the particle starts from rest at r (4, 3) m? 2. A block slides on a rough horizontal surface from point A to point B. A force of magnitude P- 2.0 N acts on the block between A and B, as shown. Points A and B are 1.5 m apart. The kinetic energies of the block at A and B are 5.0 J and 4.0 J, respectively. a) How much work is done on the block by the force P? b) How much work is done on the block by the force of friction fip 40°Explanation / Answer
(1) ans
According to the concept of the motion in one dimensional
Given that
mass m=0.5 kg
force F=x^2i+4y j=4^2 i+4*3 j=16i+12 j=20 N
position S=(4^2+3^2)^1/2=5 m
now we find the speed
f=m(v^2/2s)
20=0.5*v^2/2*5
200=0.5*v^2
400=v^2
speed v=20 m/s
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(2) ans
According to the concept of the work energy and power
Given that
force F=2 N
initial kinetic energy K.E i=5 J
finial kinetic energy K.Ef=4 J
angle=40 degree
distance s=1.5 m
now we find the work done by the force
W=Fs cos(theta)=2*1.5*cos(40)=2.3 J
now we find the work done by the friction
W+Wf=K.Ef-K.Ei
2.3+Wf=4-5=-1 j
Wf=-1-2.3=-3.3 j
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