Question

Q16:- A small cbject of mass m is suspended from a sting of length 2m. The object revolves with constant speed v in a horizontal circle of radius r When 4, The speed of the object in ms-1 (A) 4.60 (B) 5.32 (C)5.95 (D) 2.66 (E) 3.76 17If it takes 4.00 J of work to stretch a Hooke's-law spring 10.0 cm from its unstressed ength, determise the extra work required to stretch it an additional 4.0 cm. (A)6, 24) (B)7. 56/ (C)2.76/ (D)3. 84) (E)5.00/ A in the following Fig.1 The track is Q18:-А 100 ki: block is released from rest at point frictionless excopt 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 Ntn, and com presses th: spring 0.300 r. from its equilibrium position before coming to rest momcntarily Determine the coefficient of kinetic friction between the block and the rough surfia ce between points B and C. (Height-3m). (A)0. 43 (B)0.16 (C)0.33 (D)0.39 (0.41 A partick: of mass m-7.00 kg is released from point A and slides on the frictionless track showwn in Fig 2 Q15-De termine the particle's speed at point B (A) 5.8ms-1 (B) 6.2ms1 (C)5.6ms1 (D)6.00ms (E)6.4ms 20-the set work done by the gravitational force in moving the par ticle from A to B (Ao (B)162 (C)144) (D)126 (E)130

Answer 1

16)

v = (rgtan@)^0.5

= (Lsin@gtan@)^0.5

= (2sin45 x 9.81 x tan45)^0.5 = 3.725 m/s

17)

W = 1/2 kx^2

W2/W1 = x2^2 / x1^2

W2/4 = 14^2/10^2

W2 = 7.84J

W2-W1 = 3.84J

18)

Using work energy principle

Wg = mgh = 10x10x 3 = 300J

Ws = 1/2 k x^2 = 1/2 x 2250 x 0.3^2 = 101.25 J

Wf = -nmgs = -nx10x10x6.3 = -630n

By balancing we get

Wg+Wf = Ws

300 - 630n = 101.25

n = 0.3155

20)

mgh = 7x10x2 = 140J

19)

v = (2gh)^0.5 = (2x10x2)^0.5

= 6.4 m/s