A small block slides along a path that is without friction until the block reaches the section l = 3m, which begins at height h=3m on a flat incline of angle 37, as shown. In that section, the coefficient of kinetic friction is 0.50. The block passes through a point a with a speed of136 m/s . Find the speed of the block as it passes through point b where the friction ends.
Answers
Answer:
Length, l = 3m
Height, h=3m
The angle of inclination, θ = 37°
Coefficient of kinetic friction, µ = 0.50
The speed of the block when it passes through point a, VA = √136 m/s
From figure attached below, we can write
AB = h/ Sinθ = 3 / sin37° …… (i)
So, we will have
Vb² = VA² – 2gS
Here S = AB sinθ
∴ Vb² = VA² – 2gS
⇒ Vb² = VA² - 2g AB * sin θ
⇒ Vb² = ( √136)² - 2 * 10 Sin 37° * ( 3/ Sin37°) ….. [from (i)]
After cancellation,
⇒ Vb² = 136 - 60
⇒ Vb = √76 m/s
Taking point B, for the block we get
mgsinθ + μmgcosθ = ma
⇒ a = g (sinθ + μcosθ)
⇒ a = 10 (sin 37° + 0.5*cos 37°)
⇒ a = 10 m/s²
Now, in order to get the speed of the block at point B where the friction ends, we will use Newton’s Law of motion, i.e.,
VB² = Vb² - 2al …… [where l = the length of the section which is given as 3 m]
⇒ VB² = (√76)² - 2 x 10 x 3
⇒ VB = √16
⇒ VB = 4 m/s
Thus, the speed of the block as it passes through point b where the friction ends is 4 m/s.
