Question

18:-A 25.0-kg block is released from rest at point A in the following Fig IHe ValXII frictionless except for the portion between points B and C, which has a length of 6.00 m. The block travels down the track, hits a spring of force constant 2250 N/m, and compresses the spring 0.300 m fro its equilibrium position before coming to rest momentarily. Determine the coef ficient of kinetic friction between the block and the rough surface between points B and C. (Height-3m). (A)0.43 (B)0. 16(C)0.33 (D)0.39 (E)0.41 A particle of mass m - 10.00 kg is released from point A and slides on the frictionless track shown in Fig.2 Q19:-Deier mine the particle's speed at pointB (A) 5.8ms-1 (B) 6.2ms1 (C) 5.6ms1 (D)6.00ms1 (E)6.4ms-1 020: the net wor k done by the gravitational force in moving the particle from A to B. (A)90) (B)1.62 (C)144) (D)126 (E)180) 6.00 m 5,00 m ll . II . TL'y 2.00 Fig.L Prob (18) Fig. 2 Prob (19.20

Answer 1

19.)

By Using energy conservation,

KEi + PEi = KEf + PEf
KEi = 0, intially speed is zero
PEi = KEf + PEf

here, PE = m*g*h

where, m = mass of particle = 10.00 kg

g = 10 m/sec^2

h = height of object from ground

hi = initial height of particle from ground = 5.00 m

hf = final height of particle from ground = 3.20 m

So, PEi = m*g*hi

PEi = 10.00*10*5.00 = 500 J

PEf = m*g*hf

PEf = 10.00*10*3.20 = 320 J

now kinetic energy is given by,

KE = 0.5*m*v^2

So, KEf = 0.5*m*vf^2

where , vf = velocity of particle at B point = ??

So,

500 = 0.5*m*vf^2 + 320

vf^2 = 180/(0.5*10.00) = 36

vf = 6 m/sec

So, correct option is (D.)

20.)

work done by gravity(W) = change in potential energy = dPE

W = PEi - PEf

W = 500 - 320

W = 180 J

therefore correct option is (E)

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