Q: A pendulum of mass 1 kg and length l=1 ~m is released from rest at angle =60^ . The power delivered by all the forces acting on the bob at angle =30^ will be ( . Take, .g=10 ~m / s ^2 )

Power of tension =0 aligned & Power of m g=(m g)(v) 60^ \\ & Here, v= 2 g h \\ & and h=l ( 30^ - 60^ )=0.36 ~m \\ & v= 2 10 0.36 \\ & =2.68 ~m / s \\ & Power =(m g)(v) 60^ \\ & =(1 10)(2.68) 1 2 \\ & =1.34 10=13.4 ~W aligned

Question 1

A force of F - 0.5 ~N is applied on lower block as shown in figure. The work done by lower block on upper block for a displacement of 3 m of the upper block with respect to ground is (Take, .g=10 ~m /

Accepted Answer

Maximum acceleration of 1 kg block may be a_ = g=1 ~m / s ^2 Common acceleration without relative motion between two blocks may be, a= 0.5 3 ~m / s ^2 Since, a a_ There will be no relative motion and blocks will move with acceleration 0.5 3 ~m / s ^2. Force of friction by lower block on upper block, aligned f & =m a=(1) ( 0.5 3 )= 1 6 ~N ( towards right ) \ W & =f s \ & = 1 6 3=0.5 ~J aligned

Question 2

A pendulum of mass 1 kg and length l=1 ~m is released from rest at angle =60^ . The power delivered by all the forces acting on the bob at angle =30^ will be ( . Take, .g=10 ~m / s ^2 )

Accepted Answer

Power of tension =0 aligned & Power of m g=(m g)(v) 60^ \ & Here, v= 2 g h \ & and h=l ( 30^ - 60^ )=0.36 ~m \ & v= 2 10 0.36 \ & =2.68 ~m / s \ & Power =(m g)(v) 60^ \ & =(1 10)(2.68) 1 2 \ & =1.34 10=13.4 ~W aligned

Question 3

An ideal massless spring S can be compressed 1 m by a force of 100 N in equilibrium. The same spring is placed at the bottom of a frictionless inclined plane inclined at 30^ to the horizontal. A 10 kg

Accepted Answer

F=k x k= F x = 100 1 ~N / m =100 ~N / m Now, from energy conservation, between natural length of spring and its maximum compression state. 1 2 m v^2+m g h= 1 2 k x_ ^2 aligned v & = k x_ ^2 m -2 g h \ & = (100)(2)^2 10 -(2)(10) ( 2 2 ) = 20 ~m / s aligned

Question 4

A spherical ball of mass 20 kg is stationary at the top of a hill of height 100 m . It rolls down a smooth surface to the ground, then climbs up another hill of height 30 m and finally rolls down to a

Accepted Answer

Using conservation of energy, m (10 100)= m ( 1 2 v ^2+10 20 ) or 1 2 v^2=800 or v= 1600 =40 ~m / s

Question 5

A particle of mass 2 kg is on a smooth horizontal table and moves in a circular path of radius 0.6 m . The height of the table from the ground is 0.8 m . If the angular speed of the particle is 12 rad

Accepted Answer

Angular momentum L_0= mvr 90^ =2 0.6 12 1 1 [ . As V=r , . Sin 90^ =1 ] So, L_0=14.4 kgm ^2 / s

Question 6

A machine gun fires 300 bullets per min if the mass of each bullet is 10 ~g and the velocity of the bullets is 600 ~ms ^ -1 , the power (in kW ) of the gun is

Accepted Answer

Work done by the gun = total kinetic energy of the bullets aligned & =n 1 2 m v^2=300 1 2 10 10^ -3 (600)^2 \ & =150 10 10^ -3 600 600 aligned aligned Power of gun & = work done Time taken \ & = 150 10 10^ -3 600 600 60 ~s \ & =9 ~kW ( 1 minute =60 seconds ) aligned

Work Power Energy — Previous Year Questions & Solutions