NEETPhysicsWork, Power and Energy
A block of mass (m ) is stationary with respect to a rough wedge as shown in figure. Starting from rest in time (t ), ( (m=1 ~kg , =30^ , a=2 ~m / s ^2, t=4 ~s )) work done on block ( array l|l Column I & Column II (A) By gravity & (p) 144 ~J (B) By normal reaction & (q) 32 ~J (C) By friction & (r) 56 ~J (D) By all the forces & (s) 48 ~J & (t) None array )
Options
- A(A p, B t, C s, D q )
- B(A t, B p, C s, D q )
- C(A s, B p, C p, D q )
- D(A q, B p, C s, D t )
Correct answer
B. (A t, B p, C s, D q )
Step-by-step solution
( aligned & In t=4 ~s , & =a t=8 ~m / s and s = 1 2 a t^2=16 ~m & KE = 1 2 m v^2=32 ~J aligned ) From work-energy theorem, Work done by all the forces (= KE =32 ~J ) ( aligned & Work done by gravity =-m g h =-(1)(10)(16) & =-160 ~J aligned ) Writing equation of motion, we have, ( F_y=m a ) ( array lrl N 30^ +f 30^ -10=m a & =2 or & 3 N+f & =24 (i) & F_x & =0 & N 30^ & =f 30^ or & N & = 3 f (ii) array ) Solving Eqs. (i) and (ii), we have (f=6 ~N ) ( aligned & and N = 18 3 =6 3 ~N & Now, W_N =(N )(s) & =(6 3 ) ( 3 2