Q: The kinetic energies of two similar cars A and B are 100 J and 225 J respectively. On applying breaks, car A stops after 1000 m and car B stops after 1500 m . If F _ A and F _ B are the forces applied

aligned & W.D. = KE \\ aligned F . S = KE aligned & ( K E)_A ( K E)_B = - F _A S_A - F _ B ~S _ B \\ & -100 -225 = - F _ A (1000) - F _ B (1500) \\ & F _ A ~F _ B = 2 3 aligned

Q: At any instant of time t, the displacement of any particle is given by 2 t-1 (SI unit) under the influence of force of 5 N . The value of instantaneous power is (in SI unit):

aligned x & =2 t-1 \\ v & = d x d t =2 ~m ~s ^ -1 \\ P & =F . v \\ & =2 5=10 ~W aligned

Question 1

A particle of mass M moves along a horizontal x axis from x=0 to x=L. The coefficient of kinetic friction varies as a function of x as _k(x)= _0- x, where _0, are constants of appropriate dimensions,

Accepted Answer

Given the coefficient of kinetic friction _k(x) = _0 - x At x = L, _k(L) = 0 _0 - L = 0 = _0 L The frictional force acting on the particle is f_k = _k(x) Mg = ( _0 - _0 L x )Mg The magnitude of work done by the frictional force is given by: |W| = _ 0 ^ L f_k dx |W| = _ 0 ^ L ( _0 - _0 L x ) Mg dx |W| = Mg [ _0 x - _0 x^2 2L ]_ 0 ^ L |W| = Mg ( _0 L - _0 L^2 2L ) = 1 2 _0 MgL Comparing this with n _0 MgL, we get n = 1 2 Answer: 1 2

Question 2

The power of a crane, which lifts a mass of 1000 kg to a height of 20 m in 10 s is : (g = 9.8 m/s^2)

Accepted Answer

Given m = 1000 kg, h = 20 m, t = 10 s, g = 9.8 m/s^2. Work done by the crane is W = mgh. W = 1000 9.8 20 = 196000 J. Power is the rate of doing work, P = W t . P = 196000 10 = 19600 W. P = 19.6 kW. Answer: 19.6 kW

Question 3

A bob of heavy mass m is suspended by a light string of length l. The bob is given a horizontal velocity v _0 as shown in figure. If the string gets slack at some point P making an angle from the hori

Accepted Answer

C.O.M.E. 1 2 mv _0^2= mg (1+ )+ 1 2 mv _ p ^2....(1) At pt P Tp + mg = mv _ p ^2 ( as T _ p =0 ) mg = mv _ p ^2 mv _ p ^2= mg .( ii ) From (i) & (ii) aligned & 1 2 m v_0^2=m g (1+ )+ 1 2 m g \ & v_0^2=2 g (1+ )+g \ & v_0= 2 g +3 g ....(iii) aligned aligned & v _ p = g \ & v _0= 2 ~g +3 ~g \ & v _ p v _0 = 2+3 aligned

Question 4

The kinetic energies of two similar cars A and B are 100 J and 225 J respectively. On applying breaks, car A stops after 1000 m and car B stops after 1500 m . If F _ A and F _ B are the forces applied

Accepted Answer

aligned & W.D. = KE \ aligned F . S = KE aligned & ( K E)_A ( K E)_B = - F _A S_A - F _ B ~S _ B \ & -100 -225 = - F _ A (1000) - F _ B (1500) \ & F _ A ~F _ B = 2 3 aligned

Question 5

An object moving along horizontal x-direction with kinetic energy 10 J is displaced through x=(3 i ) m by the force F =(-2 i +3 j ) N . The kinetic energy of the object at the end of the displacement

Accepted Answer

Work energy theorem, W_ al = K . E (W_ al = . work done by all forces ) K_f-K_l= F x K_f-10=(-2 i +3 j ) (3 i ) K_f-10=-6 K_ f =4 ~J

Question 6

An object falls from a height of 10 m above the ground. After striking the ground it loses 50 \% of its kinetic energy. The height upto which the object can rebounce from the ground is:

Accepted Answer

K.E. just before striking the ground K_1=m g h_1=m g(10) K.E. just after striking the ground =K_1- 50 100 K_1= 50 100 K_1= K_1 2 =K_2 Now, K_2=m g h_2 K_1 2 =m g h_2 m g(10) 2 =m g h_2 h_2=5 ~m

Question 7

At any instant of time t, the displacement of any particle is given by 2 t-1 (SI unit) under the influence of force of 5 N . The value of instantaneous power is (in SI unit):

Accepted Answer

aligned x & =2 t-1 \ v & = d x d t =2 ~m ~s ^ -1 \ P & =F . v \ & =2 5=10 ~W aligned

Work Power Energy — Previous Year Questions & Solutions