Question

a wooden slab starting from rest slide down o inclined plane length 10 M with an acceleration of 5 metre per second square what would be its speed at the bottom of the inclined plane

Answer 1

Given,
s = 10m
a = 10m/s²
u = 0 [∵ starts sliding from rest]

Using, v² = u² + 2as
v² = 0 + 2(5)(10)
v² = 100 ⇒ v = 10m/s

The speed at the bottom of inclined plane is 10m/s

Answer 2

$\huge\mathfrak{Bonjour!!}$

$\huge\bold\pink{ANSWER:}$

Here, u= 0, s= 10m, a= 5m/s²

Using the 3rd equation of motion, v² - u² = 2as, we get,

→ v² - 0² = 2 × 5 × 10 = 100

or

v= √100 = 10 m/s

Therefore, the speed of the slab at the bottom of inclined plane= 10 m/s.

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