A 2 kg block is pushed up an incline plane of inclination θ = 370 imparting it a speed of 20 m s–1. How much distance will the block travel before coming to rest? The coefficient of kinetic friction between the block and the incline plane is μk = 0.5. Take g = 10 m s–2 and use sin 370 = 0.6, cos 370 = 0.8.
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F = mgsinθ + μmgcosθ
Also , F = ma
we get ,
ma = mgsinθ + μmgcosθ
a = gsinθ + μgcosθ
a = g(sinθ + μcosθ)
a = g(sin370 + 0.5 x cos370)
a = 10 (0.6 + 0.5 x 0.8)
a = 10 (0.6 + 0.4)
a = 10 x 1 = 10 m/s²
we know ,
v² - u² = 2 as
final velocity (v) = 0 (as it finally comes on rest )
initial velocity (u) = 20m/s
acceleration (a) = 10m/s²
v² - u² = 2as
s = v² - u² / 2a
s = 0 - 20²/ 2 x 10
s = -400 / 20
s = -20 m
Also , F = ma
we get ,
ma = mgsinθ + μmgcosθ
a = gsinθ + μgcosθ
a = g(sinθ + μcosθ)
a = g(sin370 + 0.5 x cos370)
a = 10 (0.6 + 0.5 x 0.8)
a = 10 (0.6 + 0.4)
a = 10 x 1 = 10 m/s²
we know ,
v² - u² = 2 as
final velocity (v) = 0 (as it finally comes on rest )
initial velocity (u) = 20m/s
acceleration (a) = 10m/s²
v² - u² = 2as
s = v² - u² / 2a
s = 0 - 20²/ 2 x 10
s = -400 / 20
s = -20 m
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