A block of mass 4 kg is moved from rest on a smooth inclined plane of inclination 53° by applying a constant force of 40 N parallel to the incline. The force acts for two seconds.
Text Solution
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Taking F = 40 N, m = 4 kg, θ = 53º
a x = (F cos α – mg sin θ )/m
= (40 cos α – 40 × 
)/4 = 10 cos α – 8
a y = 
= 
= 10 sin α
x = 0 × 2 + 
(10 cos α – 8) (2) 2 = 20 cos α – 16
y = 
(10 sin α ) (2) 2 = 20 sin α
W F = (F cos α ) x + (F sin α ) y
W F = (40 cos α ) (20 cos α – 16) + (40 sin α ) 20 sin α
= 800 cos 2 α – 640 cos α + 800 sin 2 α
W F = 800 – 640 cos α
W F ≥ 800 – 640
W F ≥ 160 J
If ; W F = 160 J then 160 = 800 – 640 cos α ⇒ cos α = 1
y = 0 and x = 20 – 16 = 4
W G = (–mg sin θ ) (4) = (–4 × 10 × 
) 4
= –128 J
F acts along the x-axis
F, x- v
W G + W F = Δ K
–128 + 160 = Δ K, K f = 32 J.
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