Data: g = 9.8 m/s^2 F = mv^2/r; a = v^2/r; t_max = V_o/g V_av = 1/2 (v_o + v); d

Question

Data: g = 9.8 m/s^2 F = mv^2/r; a = v^2/r; t_max = V_o/g V_av = 1/2 (v_o + v); delta x = V_av t F delta t = mv_t - mv_i v_r = v_of - gt v^2 = vo^2 + 2a delta x; theta = tan^-1 (y/x); Fx = F Cos theta; Fy = F Sin theta F = m.a KE = 1/2 mV^2 W = KE_f - KE_o W = F cos theta delta x; PE = mgy; C = 2 pi r sigma F_u = ma; f = mu N y = v_o t - 1/2 g t^2 Y_max = V^2 _o/2g W = F. d A block of mass M = 120 kg is puled along a rough flow by a force F = 1020 N. The force makes an angle theta = 50 degree with horizontal and the coefficient of friction mu = 0.15. The block attains an acceleration equal to A block of mass 20 kg is initially moving with velocity 30 m/s. If 550 J of work is done on the block, calculate the final velocity attained by the block [Work - Energy theorem] A car of mass 1500 kg is initially moving at 35 m/s. When the driver sees a deer ahead he slams on the breaks stopping at a distance of 125 m and just misses hitting the deer. Use the Work-Energy Theorem and determine the force necessary to bring the car to a stop. A stone of mass 90 g is whirled m a circle along a rope of length 2.8 m, if the stone makes 300 revolutions per minute, the centripetal force on the stone is 80g of ice whose temperature is - 20 degree C is to e heated until it turns to water and boils. The specific heat of is is 0 5 cal/g degree C and the heat of fusion of ice is 80 cal/g. The specific heat of water is 10 cal/g degree C and water boils at 100 degree C

Explanation / Answer

1) The normal force N on this block is due accleration due to gravity therefore, N=mg=120*9.81N=1177.2N

coefficient of friction = .15 then frictional force= coefficient of friction *N=.15*1177.2 Newton= 176.58Newton

applied force =1020N which is done via pulling the block at an angle 50degree with horizontal then

net force on the block= 1020cos50-176.58N= 479.06N

total mass of the body m=120kg

then accleration a=479.06N/120kg= 3.9m/s^2

2) mass of the block m= 20kg, initial velocity u= 30m/s

W= work done=550J= F.d

again W=change in kinetic energy=1/2mv^2-1/2mu^2

using values we get 550=1/2*m(v^2-u^2)=1/2*20(v^2- 30^2)

or 550*2/20=v^2- 900

or v=sqrt(955)=30.90m/s

3) m=1500kg, initial velocity u=35m/s

d=125m, F=?

W=change in kinetic energy=1/2mv^2-1/2mu^2

again W=F.d

then F.d=1/2mv^2-1/2mu^2

or F.125=1/2* 1500(0^2-35^2)

or F=[1500/2 *35^2]/125 N= 7350N

4) stone mass m= 90g, rope length L=2.8m

Circumferance of the rope=2*3.14*2.8/2=8.8m

stone mkes 300 rev per min

then it makes 300rev per 60sec

then in one sec no of revolution=300/60=2

we get 2 revolution in 1 sec

then 1 revolution in 1/2 sec

it travels 8.8m in 1/2 (.5) sec

then tangential velocity v=8.8/.5m/s=17.6m/s

then cetripetal force F=mv^2/r=.09*17.6^2/1.4 N=19.91N

5) Required energy=mst+mL

=80*.5(100-(-20)+80*80=4800+6400=11200 calorie

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