Work Power And Energy(11th And 12th > Physics ) Questions and Answers
Explanation:-
Answer: Option A. -> 12kx2:
A
Total work done is equal to the energy stored in the spring.
A block of mass 5.0 kg slides down from the top of an inclined plane of length 3 m. The first 1 m of the plane is smooth and the next 2 m is rough. The block is released from rest and again comes to rest at the bottom of the plane. If the plane is inclined at 30∘ with the horizontal, find the coefficient of friction on the rough portion.
Explanation:-
Answer: Option B. -> √32:
B
Force Method:
Form P to Q: v2=O2+2a1S1
Where a1=gsin 30∘,s1=1m
And form Q to R: O2=v2+2a2s2
Where a2=gsin 30∘−μgcos 30∘,s2=2m
Solve to get μ=√32
A small block of mass 2 kg is kept on a rough inclined surface of inclination θ=30∘ fixed in a lift. The lift goes up with a uniform speed of 1 ms−1 and the block does not slide relative to the inclined surface. The work done by the force of friction on the block in a time interval of 2 s is.
Explanation:-
Answer: Option B. -> 9.8 J:
B
Since the lift's speed is constant, in the absence of acceleration, there will be no pseudo-force (referring to the lift as the frame of reference) or additional reaction/thrust due to inclined plane (referring to ground as the reference frame) on the particle.
Therefore, friction =mg sinθ acting along the plane.
Distance moved by the particle (or lift) in time t=vt
Work done in time t=(mg sinθ)vt(cos(90∘- θ))=mg sin2θ vt
Substituting the Values we get Work = 9.8 J
Explanation:-
Answer: Option C. -> mk4t28:
C
GIven v=k√x or dxdt=k√x or x−12dx=kdt
Integrating both sides, we get
x1212=kt+C; Assuming x(0) = 0
Therefore, C = 0
2√x=kt⇒x=k2t24 or v=k2t2
Therefore, work done,
△W = Increase in KE
= 12mv2−12m(0)2=12m[k2t2]2=mk4t28
Explanation:-
Answer: Option D. -> 180∘:
D
⃗A.⃗B=|A||B|cosθ or,cosθ=⃗A.⃗B|A||B| ----------------(i)
But, ⃗A.⃗B=(^i+^j+^k).(−2^i−2^j−2^k)
⃗A.⃗B=−2−2−2
⃗A.⃗B=−2−2−2=−6
Again A = |⃗A|=√(1)2+(1)2+(1)2=√3;
B = |⃗B|=√(−2)2+(−2)2+(−2)2=√12=2√3
Now,cosθ=−6√3×2√3=−1⇒θ=180∘
Explanation:-
Answer: Option B. -> -7:
B and C
→F=3^i+2^j+c^k
→r=c^i+4^j+c^k
Work = 36j
W = →F.→r
36 = (3^i+2^j+c^k).→F=(c^i+4^j+c^k)
36 = 3c + 8 + c2
⇒c2+3c−28=0
(c+7) (c - 4)
c = -7 or 4