First slide

Work done by Diff type of forces

Question

A block of mass m is stationary with respect to a rough wedge is shown in figure. Starting from rest in
time $t (m = 1 kg, θ = 30 °, a = 2 {ms}^{- 2}, t = 4 s)$
Match the following columns for work done on the block and mark the correct option from the codes given below.
Column I
(A) By gravity
(B) By normal reaction
(C) By friction
(D) By all the forces
Column II
(p) 144 J
(q) 32 J
(r ) $-$ 160J
(s) 48 J

Difficult

A

A B C D
r q s p
B

A B C D
q r s p
C

A B C D
p q r s
D

A B C D
r p s q

Solution

In $t = 4 s, v = a t = 8 {ms}^{- 1} and s = \frac{1}{2} a t^{2} = 16 m$
$KE = \frac{1}{2} m v^{2} = 32 J$
From work-energy theorem,
Work done by all the forces $= Δ K E = 32 J$
Work done by gravity, $W_{g} = - m g h = - (1) (10) (16) = - 160 J$
Writing equation of motion, we have
$N \cos 30 ° + f \sin 30 ° - 10 = m a = 2$
or $\sqrt{3} N + f = 24$ …(i)
$Σ F_{x} = 0$
$∴ N \sin 30 ° = f \cos 30 ° or N = \sqrt{3} f$ …(ii)
Solving Eqs. (i) and (ii), we have
$f = 6 N and N = 6 \sqrt{3} N$
Now, work done by normal reaction, $W_{N} = (N \cos θ) (s)$
$= (6 \sqrt{3}) (\frac{\sqrt{3}}{2}) (16) = 144 J$
Work done by friction, $W_{f} = (f \sin θ) (s) = (6) \frac{1}{2} (16) = 48 J$
Work done by all the forces,
$W = W_{g} + W_{N} + W_{F} = - 160 + 144 + 48 = 32 J$
Hence, $A \to r, B \to p, C \to s, D \to q$

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